Category Archives: លំហាត់ស្វ៊ីត

លំហាត់ស្វ៊ីត

  • បណ្តាលំហាត់ស្វ៊ីតដកស្រង់ចេញពី​វិញ្ញា សារប្រលងអូឡាំពិច
  • 1- គេអោយស្វ៊ីត{x_{k}}កំតន់ដោយ:

    x_{k}= \dfrac{1}{2!}+ \dfrac{2}{3!}+...+ \dfrac{k}{(k+1)!}.

    គណនា \displaystyle \lim_{n \to +\infty} \sqrt[n]{x_{1}^{n}+x_{2}^{n}+...+x_{1999}^{n}}.

  • សំរាយ
  • ដោយ x_{k+1}-x_{k}= \dfrac{k+1}{(k+2)!} > 0, \forall k \in N \Rightarrow x_{k+1} > x_{k} > 0, \forall k \in N

    \Rightarrow x_{1999}^{n} < x_{1}^{n}+x_{2}^{n}+...+x_{1999}^{n} < 1999.x_{1999}^{n}

    \Rightarrow x_{1999} < \sqrt[n]{x_{1}^{n}+x_{2}^{n}+...+x_{1999}^{n}} < \sqrt[n]{1999}.x_{1999}. \quad (*)

    ក្រៅពីនេះយើងមាន:

    \dfrac{k}{(k+1)!}= \dfrac{(k+1)-1}{(k+1)!}= \dfrac{1}{k!}- \dfrac{1}{(k+1)!}

    \Rightarrow x_{k}= \bigl (1- \dfrac{1}{2!} \bigr )+ \bigl ( \dfrac{1}{2!}- \dfrac{1}{3!} \bigr )+...+ \bigl ( \dfrac{1}{k!}- \dfrac{1}{(k+1)!} \bigr ) =1- \dfrac{1}{(k+1)!}

    \Rightarrow x_{1999}=1- \dfrac{1}{2000!}

    មកដល់ទីនេះយើងជំនួសx_{1999}ចូល(*)គេបាន:

    1- \dfrac{1}{2000!} < \sqrt[n]{x_{1}^{n}+x_{2}^{n}+...+x_{1999}^{n}} < \sqrt[n]{1999} \bigl (1- \dfrac{1}{2000!} \bigr )

    តែដោយ\displaystyle \lim_{n \to +\infty} \bigl (1- \dfrac{1}{2000!} \bigr )= \lim_{n \to +\infty} \bigl [ \sqrt[n]{1999} (1- \dfrac{1}{2000!}) \bigr ]

    ដូចនេះគេបាន:

    \displaystyle \lim_{n \to +\infty} \sqrt[n]{x_{1}^{n}+x_{2}^{n}+...+x_{1999}^{n}}=1- \dfrac{1}{2000!}.

    2-គេអោយស្វ៊ីតចំនួនពិតវិជ្ជមាន {a_{n}}ផ្ទៀងផ្ទាត់វិសមភាព:

    a_{n}^{ \frac{2000}{1999}} \geq a_{n-1}+a_{n-2}+...+a_{1}, \forall n \geq 2.

    បង្ហាញថា:\exists c > 0: a_{n} > nc, \forall n \in N.

  • សំរាយ
  • យើងមាន: a_{n} > a_{1}^{ \frac{1999}{2000}} , \forall n \geq 2

    \Rightarrow a_{n}^{ \frac{2000}{1999}} > (n-2)a_{1}^{ \frac{1999}{2000}}+a_{1}, \forall n \geq 2

    \Rightarrow \displaystyle \lim_{n \to +\infty} a_{n}= +\infty

    \Rightarrow \exists n_{o} \in N: a_{n} \geq 1, \forall n \geq n_{o}

    តាង c=Min \begin{Bmatrix} \dfrac{1}{4},a_{1},  \dfrac{a_{2}}{2},..., \dfrac{a_{n_{o}}}{n_{o}} \end{Bmatrix}

    \Rightarrow a_{n} \geq nc, \forall n \in \begin{Bmatrix} 1,2,...,n_{o} \end{Bmatrix}

    ឧបមាថា a_{n} \geq n.c, \forall n \leq n_{1} ,(n_{1} \in N,n_{1} \geq n_{o})

    យើងមាន:a_{n+1}^{2} \geq a_{n+1}^{ \frac{1999}{2000}} \geq a_{n}+a_{n-1}+...+a_{1} \geq [n+(n-1)+...+1].c

    = \dfrac{n(n+1)}{2}.c \geq (n+1)^{2}.c^{2}(ព្រោះ \dfrac{n}{2(n+1)} \geq \dfrac{1}{4} \geq c).

    \Rightarrow a_{n+1} \geq (n+1).c \Rightarrow a_{n} \geq n.c

    3- គេអោយស្វ៊ីតចំនួនពិត{a_{n}}កំន់តដោយ:

    \left\{ \begin{array}{l} a_{o}=1999 \\ a_{n+1}= \dfrac{a_{n}^{2}}{1+a_{n}}, \forall n \geq 0 \end{array} \right.

    ចូររកផ្នែកគត់របស់ a_{n}, (0 \leq n \leq 999).

  • សំរាយ
  • ពិតណាស់ថា a_{n} > 0, \forall n \geq 0, គេបាន:

    a_{n}-a_{n+1}=a_{n}- \dfrac{a_{n}^{2}}{1+a_{n}}= \dfrac{a_{n}}{1+a_{n}} > 0, \forall n \geq 0

    \Rightarrow \begin{Bmatrix} a_{n} \end{Bmatrix}ជាស្វ៊ីតចុះ (1)

    \Rightarrow a_{n+1}= \dfrac{a_{n}^{2}}{1+a_{n}}=a_{n}- \dfrac{a_{n}}{1+a_{n}} > a_{n}-1, \forall n \geq 0

    \Rightarrow a_{n+1} > a_{o}-(n+1), \forall n \geq 0

    \Rightarrow a_{n-1} > a_{o}-(n-1) ,\forall n \geq 2

    \Rightarrow a_{n-1} > 2000-n, \forall n \geq 2 \quad (2)

    ក្រៅពីនេះយើងមាន:

    a_{n}=a_{o}+(a_{1}-a_{o})+(a_{2}-a_{1})+...+(a_{n}-a_{n-1})

    \quad =1999- \biggl ( \dfrac{1}{1+a_{o}}+ \dfrac{1}{1+a_{1}}+...+ \dfrac{1}{1+a_{n-1}} \biggr )

    \quad =1999-n+ \biggr ( \dfrac{1}{1+a_{o}}+ \dfrac{1}{1+a_{1}}+...+ \dfrac{1}{1+a_{n-1}} \biggr ) \quad (3)

    ពី(1)និង(2)យើងមាន:

    0 < \dfrac{1}{1+a_{o}}+ \dfrac{1}{1+a_{1}}+...+ \dfrac{1}{1+a_{n-1}} < \dfrac{n}{1+a_{n-1}}

    < \dfrac{n}{2001 -n} < \dfrac{n}{1998 -n} \leq 1, (2 \leq n \leq 999)​ ,(4)

    ពី(3)និង(4)យើងមាន:

    1999-n < a_{n} < 1999-n+1, \quad (2 \leq n \leq 999)

    \Rightarrow [a_{n}]=1999-n, \quad (2 \leq n \leq 999).

    យើងពិនិត្យដោយផ្ទាល់:

    +a_{o}=1999 \Rightarrow [a_{o}]=1999

    +a_{1}= \dfrac{a_{o}^{2}}{1+a_{o}}=a_{o}- \dfrac{a_{o}}{1+a_{o}}=1999- \dfrac{1999}{2000}

    \Rightarrow a_{1}=1998+ \dfrac{1}{2000} \Rightarrow [a_{1}]=1998

    ដូចនេះ[a_{n}]=1999-n, \quad (0 \leq n \leq 999).

    4- គេអោយស្វ៊ីតចំនួនពិត\begin{Bmatrix} a_{n} \end{Bmatrix}, \begin{Bmatrix} b_{n} \end{Bmatrix}ផ្ទៀងផ្ទាល់:

    \left\{ \begin{array}{l} a_{1}=3 \\ b_{1}=2 \\ a_{n+1}=a_{n^{2}}+2b_{n^{2}} \\ b_{n+1}=2a_{n}b_{n} \end{array} \right. \quad ( \forall n \in N)

    a)ស្រាយបំភ្លឺថា a_{n}, b_{n} ជាពីរចំនួនបថមនិងគ្នា។
    b)រករូបមន្តអោយ a_{n}, b_{n}.

  • សំរាយ
  • a)ប្រើវិចារកំនើនមានកំនត់:

    +n=1 \Rightarrow a_{1}^{2}-2b_{1}^{2}=3^{2}-2.2^{2}=1.

    ឧបមាថាពិតដល់ k គឺថាa_{k}^{2}-2b_{k}^{2}=1.
    យើងនឹងស្រាយថាពិតដល់ k+1.យើងមាន:

    a_{k+1}^{2}-2b_{k+1}^{2}=(a_{k}^{2}+2b_{k}^{2})^{2}-2(2a_{k}b_{k})^{2}=(a_{k}^{2}-2b_{k}^{2})^{2}=1.

    នោះ a_{n}^{2}-2b_{n}^{2}=1, \quad n \in N. \quad (1)

    តាង d ជាផលចែករួមធំបំផុតរបស់ a_{n},b_{n} នោះ d​​ ក៏ជាផលចែករួមរបស់ a_{n}^{2}និង b_{n}^{2}។នាំអោយ d គឺជាផលចែករួមរបស់ a_{n}^{2}-2b_{n}^{2}=1,ហេតុនោះ d=1.

    ដូចនេះ a_{n},b_{n}ជាពីចំនួនបថមនិងគ្នា។

    b)យើងមាន:
    a_{n+1}-b_{n+1} \sqrt{2}=a_{n}^{2}+2b_{n}^{2}-(2a_{n}b_{n}) \sqrt{2}=(a_{n}-b_{n} \sqrt{2})^{2}

    \Rightarrow lg(a_{n+1}-b_{n+1} \sqrt{2})=2lg(a_{n}-b_{n} \sqrt{2})

    \Rightarrow lg(a_{n+1}-b_{n+1} \sqrt{2}=2^{n}lg(a_{1}-b_{1} \sqrt{2})=2^{n}lg(3-2 \sqrt{2})=2^{n+1}lg( \sqrt{2}-1)

    \Rightarrow a_{n+1}-b_{n+1} \sqrt{2}=( \sqrt{2}-1)^{2^{n+1}} \quad (2)

    តាម(1)និង(2)គេបាន:\left\{ \begin{array}{l} a_{n}+b_{n} \sqrt{2}=( \sqrt{2}+1)^{2^{n}} \\ a_{n}-b_{n} \sqrt{2}=( \sqrt{2}-1)^{2^{n}} \end{array} \right.

    ដូចនេះ \left\{ \begin{array}{l} a_{n}= \dfrac{1}{2}[( \sqrt{2}+1)^{2^{n}}+( \sqrt{2}-1)^{2^{n}}] \\ b_{n}= \dfrac{1}{2 \sqrt{2}}[( \sqrt{2}+1)^{2^{n}}-( \sqrt{2}-1)^{2^{n}}] \end{array} \right.

    5- គេអោយស្វ៊ីត{a_{n}}ផ្ទៀងផ្ទាល់:

    a_{o}=1, a_{1000}=0, a_{n+1}=2a_{1}a_{n}-a_{n-1}, (n \geq 1).

    គណនាផលបូក:a_{1999}+a_{1}.

  • សំរាយ
  • ដំបូងយើងត្រូវបង្ហាញថា |a_{1}| < 1.
    ប្រើវិធីផ្ទុញគឺឧបមា:|a_{1}| > 1.យើងមាន:

    |a_{2}|=|2a_{1}^{2}-a_{o}|=2a_{1}^{2}-1 \geq |a_{1}|
    តាមវិចារកំនើនមានកំនត់យើងមាន:

    |a_{n+1}|= |2a_{1}a_{n}-a_{n-1}|

    \geq 2|a_{1}|.|a_{n}|-|a_{n-1}|

    \geq 2|a_{n}|-|a_{n-1}| \geq |a_{n}|

    នោះ |a_{n}| \geq |a_{n-1}| \geq ... \geq |a_{o}|=1.

    រើស n=1000, \Rightarrow |a_{1000}| \geq 1(ផ្ទុយពីសម្មតិកម្ម)។
    ហេតុនោះ |a_{1}| < 1.

    តាង :a_{1}=cos \varphi, \quad \varphi \in (o, \pi )

    \Rightarrow a_{2}=2a_{1}^{2}-1=2cos \varphi^{2}-1=cos2 \varphi.

    តាមវិចារកំនើនមានកំនត់យើងមាន:

    a_{n+1}=2a_{1}a_{n}-a_{n-1}

    =2cos \varphi cosn \varphi-cos(n-1) \varphi

    =cos(n+1) \varphi +cos(n-1) \varphi -cos(n-1) \varphi =cos(n+1) \varphi

    ហេតុនោះ a_{1000}=cos1000 \varphi =0

    \iff 1000 \varphi = \dfrac{ \pi}{2}+k. \pi, k \in Z

    \Rightarrow a_{1999}=cos1999 \varphi =cos(2000 \varphi - \varphi )=cos( \pi +2k \pi - \varphi )=-cos \varphi =-a_{1}.

    ដូចនេះ:a_{1999}+a_{1}=0.

    6- តើមានប៉ុន្មានស្វ៊ីតចំនួនពិតវិជ្ជមាន {a_{n}}ដែលផ្ទៀងផ្ទាល់:

    a_{o}=1,a_{1}=2,|a_{n+2}.a_{n}-a_{n+1}^{2}|=1?

  • សំរាយ
  • យើងមានដ្យាក្រាមកំនត់ដូចខាងក្រោម:

    a_{o}=1, a_{1}=2, a_{2}= \begin{bmatrix} 3 \Rightarrow a_{3}= \begin{bmatrix} 5 \\ 4 \end{bmatrix} \\ 5 \Rightarrow a_{3}= \begin{bmatrix} 13 \\ 12 \end{bmatrix} \end{bmatrix}

    យើងនឹងស្រាយថាមានបួនស្វ៊ីតវិជ្ជមានដែលផ្ទៀងផ្ទាល់តតាមប្រធាន។
    +ស្រាយថាមានស្វ៊ីតវិជ្ជមានតែមួយគត់ដែលផ្ទៀងផ្ទាល់:
    a_{o}=1,a_{1}=2,a_{2}=3,a_{3}=5,|a_{n+1}^{2}-a_{n}a_{n+2}|=1, \forall n \geq 2

    នោះគឺជាស្វ៊ីតវិជ្ជមាន{a_{n}}ដែលផ្ទៀងផ្ទាល់:

    a_{o}=1, a_{1}=2, a_{n}=a_{n-1}+a_{n-2}, \forall n \geq 2 \quad (1)

    នោះគេបាន:

    |a_{n+1}^{2}-a_{n}a_{n+2}|=|(a_{n+1}+a_{n})a_{n}-a_{n+1}^{2}|

    =|a_{n}^{2}+a_{n+1}(a_{n}-a_{n+1})|=|a_{n}^{2}-a_{n+1}.a_{n-1}|=1, \quad \forall n \in N.

    -ប្រើវិចារកំនើនកំនត់យើងបាន (1)ជាស្វ៊ីតកើន។

    ហេតុនោះ: |a_{n+1}^{2}-a_{n}a_{n+2}|=1 \Rightarrow a_{n+2}= \dfrac{a_{n+1}^{2}+1}{a_{n}}

    ពីវិចារកំនើនមានកំនត់: a_{n+1} > a_{n} \Rightarrow a_{n} \leq a_{n+1}-1

    \Rightarrow a_{n+2} \geq \dfrac{a_{n+1}^{2} \pm 1}{a_{n+1}-1} \geq \dfrac{a_{n+1}-1}{a_{n+1}-1}=a_{n+1}+1

    \Rightarrow a_{n+2} > a_{n+1}.

    -ស្វ៊ីត(1)ត្រូវបានកំនត់មានតែមានគត់។
    ហេតុនោះ ឧបមាថាមាន​n \geq 2: a_{n}, a_{n+1} តែមួយគត់ដែលាមានពីរតំលៃ a_{n+2},a'_{n+2}, \quad a_{n+2} > a'_{n+2}ដែលផ្ទៀងផ្ទាល់របៀបកំន់តស្វ៊ីត​ គឺថា:

    \left\{ \begin{array}{l} a_{n}.a_{n+1}=a_{n+1}^{2}+1 \\ a_{n}.a_{n+2}=a_{n+1}^{2}-1 \end{array} \right. \Rightarrow a_{n}(a_{n+2}-a'_{n+2})=2 \vdots a_{n}

    \Rightarrow មិនពិត(ព្រោះ a_{n} \geq a_{2}=3 > 2)

    សរុបមកយើងស្រាយបានថាមានស្វ៊ីតវិជ្ជមានតែមួយគត់:

    a_{o}=1,a_{1},a_{2}=3,a_{3}=5, \quad |a_{n+1}^{2}-a_{n}a_{n+2}|=1, \forall n \geq 3

    នោះក៏ជាស្វ៊ីត :

    a_{o}=1, a_{1}=2, a_{n}=a_{n-1}+a_{n-2}, \forall n \geq 2

    ស្រាយដូចគ្នា យើងនឹងបង្ហាញបានថាមានបណ្តាស្វ៊ីតវិជ្ជមាន:

    a_{o}=1, a_{1}=2,a_{2}=3, a_{3}4, |a_{n+1}^{2}-a_{n}a_{n+2}|=1, \forall n \geq 2.

    a_{o}=1, a_{1}=2, a_{2}=5, a_{3}=12, |a_{n+1}^{2}-a_{n}a_{n+2}|=1, \forall n \geq 2.

    a_{o}, a_{1}=2, a_{2}=5, a_{3}=13, |a_{n+1}^{2}-a_{n}a_{n+2}|=1, \forall n \geq 2.

    នោះក៏ជាបណ្តាស្វ៊ីត(រៀងគ្នា):

    a_{o}=1, a_{1}=2,a_{n+2}=2_{n+1}-a_{n}, \forall n \in N. \quad (2)

    a_{o}=1, a_{1}=2, a_{n+2}=2a_{n+1}+a_{n}, \forall n \in N. \quad (3)

    a_{o}=1, a_{1}=2, a_{n+2}=3a_{n+1}-a_{n}, \forall n \in N. \quad (4)

    ដូចនេះ: មានស្វ៊ីតវិជ្ជមានចំនួនបួនគឺ(1),(2),(3),និង(4)ដែលផ្ទៀងផ្ទាល់តាមប្រធាន។

    7- គេអោយស្វ៊ីតចំនួនពិត{S_{n}} ជាមួយ​ S_{n}= \dfrac{n+1}{2^{n+1}} \displaystyle \sum_{k=1}^{n} \dfrac{2^{k}}{k}.

    បង្ហាញថាមាន \displaystyle \lim_{n \to +\infty} S_{n} ហើយគណនាលីមីតនោះ។

  • សំរាយ
  • យើងមាន : S_{n+1}= \dfrac{n+2}{2^{n+2}} \displaystyle \sum_{k=1}^{n+1} \dfrac{2^{k}}{k}= \dfrac{n+2}{2^{n+2}} \biggl ( \dfrac{2^{1}}{1}+ \dfrac{2^{2}}{2}+...+ \dfrac{2^{n+1}}{n+1} \biggr )

    = \dfrac{n+2}{2(n+1)}. \dfrac{n+1}{2^{n+1}} \biggl ( \dfrac{2^{1}}{1}+ \dfrac{2^{2}}{2}+...+ \dfrac{2^{n}}{n} \biggr )+ \dfrac{n+2}{2(n+1)}

    = \dfrac{n+2}{2(n+1)}(S_{n}+1)

    ស្រដៀងគ្នាដែរគេបាន:

    S_{n+2}= \dfrac{n+3}{2(n+2)}(S_{n+1}+1)

    \Rightarrow S_{n+2}-S_{n+1}= \dfrac{(n+1)(n+3)(S_{n+1}+1)-(n+2)^{2}(S_{n}+1)}{2(n+1)(n+2)}

    = \dfrac{(n^{2}+4n+3)(S_{n+1}-S_{n})-S_{n}-1}{2(n+1)(n+2)}

    នោះស្វ៊ីត {S_{n}} គឺជាស្វ៊ីតចុះ។
    ហេតុនោះមាន \displaystyle \lim_{n \to +\infty} S_{n}.តាង S​ជាលីមីតនោះ.

    ពី S_{n+1}= \dfrac{n+2}{2(n+1)}(S_{n}+1)

    \Rightarrow S= \dfrac{1}{2}(S+1) \iff S=1.

    ដូចនេះ:\displaystyle \lim_{n \to +\infty} S_{n}=1.

    8- គេដឹងវិសមភាព:

    x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2} \geq (x_{1}+x_{2}+...+x_{n-1})x_{n}.

    ផ្ទៀងផ្ទាត់ជាមួយគ្រប់ចំនួនពិត​ x_{1}, x_{2},...,x_{n}, (n \geq 1).តើ n ត្រូវស្មើរប៉ុន្មាន?

  • សំរាយ
  • ឧបមាថាវិសមភាព: x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2} \geq (x_{1}+x_{2}+...+x_{n-1})x_{n}. \quad (1)

    ផ្ទៀងផ្ទាត់ជាមួយគ្រប់ចំនួនពិត x_{1}, x_{2},...,x_{n}, (n \geq 1)

    ពេលនោះវាក៏កើតឡើងជាមួយ: \left\{ \begin{array}{l} x_{1}=x_{2}=...=x_{n-1}=1 \\ x_{n}=2 \end{array} \right.

    \Rightarrow (n-1)+4 \geq (n-1)2

    \Rightarrow 1 \leq n \leq 5

    ច្រាសមកវិញ​ ឧបមា​ 1 \leq n \leq 5, ​​ យើងនឹងស្រាយថា(1) ត្រូវបានផ្ទៀងផ្ទាត់ជាមួយ គ្រប់ចំនួនពិត​ x_{1}, x_{2},...,x_{n}.

    ហេតុនេះ យើងពិនិត្យត្រីធា:

    f(x_{n})=x_{n}^{2}-(x_{1}+x_{2}+...+x_{n-1})x_{n}+x_{1}^{2}+x_{2}^{2}+...+x_{n-1}^{2}

    នេះគឺជាត្រីធាលំដាប់ពីរ ចំពោះ x_{n} ហើយយើងមាន:

    \Delta =(x_{1}+x_{2}+...+x_{n-1})^{2}-4(x_{1}^{2}+x_{2}^{2}+...+x_{n-1}^{2})

    តាមវិសមភាព Bunhiacopski:

    4(x_{1}^{2}+x_{2}^{2}+...+x_{n-1}^{2}) \geq (n-1)(x_{1}^{2}+x_{2}^{2}+...+x_{n-1}^{2}) \geq (x_{1}+x_{2}+...+x_{n-1})^{2}

    \Rightarrow \Delta \leq 0

    \Rightarrow f(x_{n}) \geq 0, \forall x_{n} \in R.

    ដូចនេះ:n \in \begin{Bmatrix} 1, 2, 3, 4, 5 \end{Bmatrix}.

    9- កំណត់តួទូទៅរបស់ស្វ៊ីតចំនួនពិត{u_{n}},គេដឹងថា:

    \left\{ \begin{array}{l} u_{1}=2 \\ u_{n+1}=9u_{n}^{3}+3u_{n}, (n=1,2,3,...). \end{array} \right.

  • សំរាយ
  • តាង:V_{n}=3u_{n},(n=1,2,...).យើងមាន \left\{ \begin{array}{l} V_{1}=6 \\ V_{n+1}=V_{n}^{2}+3V_{n} \end{array} \right.

    រើសយក x_{1},x_{2}: \left\{ \begin{array}{l} x_{1}+x_{2}=6 \\ x_{1}.x_{2}=-1 \end{array} \right.

    +ជាមួយ n=1,យើងមាន:

    V_{1}=6=x_{1}+x_{2}=x_{1}^{3^{1-1}}+x_{2}^{3^{1-1}}

    +ជាមួយ n=k,(k \in N) យើងឧបមាថា :V_{k}=x_{1}^{3^{k-1}}+x_{2}^{3^{k-1}}.

    +ជាមួយ n=k+1, យើងមាន:

    V_{k+1}=V_{k}^{3}+3V_{k}

    = \bigl (x_{1}^{3^{k-1}}+x_{2}{3^{k-1}} \bigr )^{3}+3 \bigl (x_{1}^{3^{k-1}}+x_{2}^{3^{k-1}} \bigr )

    =x_{1}^{3^{k}}+x_{2}^{3^{k}}+3(x_{1}x_{2})^{3^{k-1}} \bigl (x_{1}^{3^{k-1}}+x_{2}^{3^{k-1}} \bigr )+3 \bigl (x_{1}^{3^{k-1}}+x_{2}^{3^{k-1}} \bigr ).

    =x_{1}^{3^{k}}+x_{2}^{3^{k}},(ព្រោះ (x_{1}x_{2})^{3^{k-1}}=(-1)^{3^{k-1}}=-1)

    នោះតាមវិចារកំនើនមានកំនត់គេបាន:

    V_{n}=x_{1}^{3^{n-1}}+x_{2}^{3^{n-1}}, \forall n \in N

    ដូចនេះ: u_{n}= \dfrac{1}{3} \bigl [(3- \sqrt{10})^{3^{n-1}}+(3+ \sqrt{10})^{3^{n-1}} \bigr ].

    (ព្រោះ x_{1},x_{2} ជារឹសរបស់សមីការ x^{2}-6x-1=0)

    10- គេអោយស្វ៊ីត {x_{n}} កំនត់ដូចខាងក្រោម:

    \left\{ \begin{array}{l} x_{1}=1 \\ x_{n+1}= \begin{bmatrix} \dfrac{3}{2}x_{n} \end{bmatrix} \quad \forall n \geq 1 \end{array} \right.

    បង្ហាញថាស្វ៊ីត x_{n}} មានចំនួនគូ និងចំនួនសេសរាប់មិនអស់។(តាង [x]​ ជាផ្នែកគត់របស់ x).

  • សំរាយ
  • +ឧបមាថាស្វ៊ីត \begin{Bmatrix} x_{n} \end{Bmatrix}^{ \alpha} គ្រាន់តែមានចំនួនគូកំនត់,​ នាំអោយមានយ៉ាងតិចចំនួន​ n \in Nមួយដែល x_{k}សេស \forall k \geq n.

    តាង :x_{k}=2^{ \alpha}. \beta +1(ជាមួយ \alpha , \beta \in N, \betaសេស.

    នោះគេបាន:

    x_{k+1}=2^{ \alpha -1}. 3 \beta +1

    x_{k+2}=2^{ \alpha -2}.3^{2} \beta +1
    ………………
    x_{k+ \alpha }=3^{ \alpha } \beta+1

    \Rightarrow x_{k+ \alpha }គឺជាចំនួនគូ​ \Rightarrow មិនពិតតាមឧបមា។

    នោះយើងទាញបានថាស្វ៊ីតខាងលើមានចំនួនគូមិនកំនត់។

    +ឧបមាថាស្វ៊ីត \begin{Bmatrix} x_{n} \end{Bmatrix}^{ \alpha } មានចំនួនសេសកំនត់, នាំអោយមានយ៉ាងតិចចំនួនn \in Nមួយដែល​ x_{k}គូ, \forall k \geq n.

    តាង :x_{k}=2^{ \alpha }. \beta, ( \alpha , \beta \in N, \quad \betaសេស)

    នោះគេបាន:

    x_{k+1}=3.2^{ \alpha -1} \beta

    x_{k+2}=3^{2}. 2^{ \alpha -2} \beta
    ………………….
    x_{k+ \alpha }=3^{ \alpha }. \beta

    \Rightarrow x_{k+ \alpha }ជាចំនួនសេស \Rightarrowមិន​ពិតតាមឧបមា។

    នោះនាំអោយស្វ៊ីតខាងលើមានចំនួនសេសរាប់មិនអស់។

    11- អោយ n ជាចំនួនពិតវិជ្ជមាន,a_{1},a_{2},...,a_{n} > 0,(n \geq 2)ផ្ទៀងផ្ទាត់:

    a_{1}+a_{2}+...+a_{n}=1

    បង្ហាញថា :\displaystyle \sum_{i=1}^{n} \sqrt{ \dfrac{1-a_{i}}{a_{i}}} \geq (n-1) \sum_{i=1}^{n} \sqrt{ \dfrac{a_{i}}{1-a_{i}}}

  • សំរាយ
  • យើងមាន:

    f= \displaystyle \sum_{i=1}^{n} \sqrt{ \dfrac{1-a_{i}}{a_{i}}}-(n-1) \sum_{i=1}^{n} \sqrt{ \dfrac{a_{i}}{1-a_{i}}}

    =\biggl ( \displaystyle \sum_{j=1}^{n}a_{j} \biggr ) \biggl ( \sum_{i=1}^{n} \sqrt{ \dfrac{1-a_{i}}{a_{i}}} \biggr )- \biggl ( \sum_{j=1}^{n}(1-a_{j}) \biggr ). \biggl ( \sum_{i=1}^{n} \sqrt{ \dfrac{a_{i}}{1-a_{i}}} \biggr )

    = \displaystyle \sum_{i=1}^{n} \sum_{j=1}^{n} \biggl ( a_{j} \sqrt{ \dfrac{1-a_{i}}{a_{i}}}-(1-a_{i}) \sqrt{ \dfrac{a_{i}}{1-a_{i}}} \biggr )

    = \displaystyle \sum_{i=1}^{n} \sum_{j=1}^{n} \dfrac{a_{j}(1-a_{i})-(1-a_{j})a_{i}}{ \sqrt{a_{i}(1-a_{i})}}= \sum_{i=1}^{n} \sum_{j=1}^{n} \dfrac{a_{j}-a_{i}}{ \sqrt{a_{i}(1-a_{i})}}

    = \displaystyle \sum_{1 \leq i < j \leq n} \biggl ( \dfrac{a_{j}-a_{i}}{ \sqrt{a_{i}(1-a_{i}}}+ \dfrac{a_{i}-a_{j}}{ \sqrt{a_{j}(1-a_{j})}} \biggr ) \geq 0

    (ព្រោះ \dfrac{a_{j}-a_{i}}{ \sqrt{a_{i}(1-a_{i})}}+ \dfrac{a_{i}-a_{j}}{ \sqrt{a_{j}(1-a_{j})}}= \dfrac{(a_{j}-a_{i})[ \sqrt{a_{j}(1-a_{j})}- \sqrt{a_{i}(1-a_{i})}]}{ \sqrt{a_{i}a_{j}(1-a_{i})(1-a_{j})}}

    = \dfrac{(a_{i}-a_{j})^{2}(1-a_{i}-a_{j})}{ \sqrt{a_{i}a_{j}(1-a_{i})(1-a_{j})}[ \sqrt{a_{j}(1-a_{j})}+ \sqrt{a_{i}(1-a_{i})}]} \geq 0.)

    12- គេអោយស្វ៊ីត {x_{n}} ជាមួយ x_{1}=a \neq -2, x_{n+1}= \dfrac{3 \sqrt{2x_{n}^{2}+2}-2}{2x_{n}+ \sqrt{2x_{n}^{2}+2}}, \forall n \in N.

    ពិនិត្យភាពបង្រួមរបស់ស្វ៊ីត និងរកលីមីតរបស់វា(បើមាន) ដោយយោងតាមករណី a.

  • សំរាយ
  • *តាង: f(x)= \dfrac{3 \sqrt{2x^{2}+2}-2}{2x+ \sqrt{2x^{2}+2}}

    និង g(x)=f(x)-x= \dfrac{-2x^{2}+(3-x) \sqrt{2x^{2}+2}-2}{2x+ \sqrt{2x^{2}+2}}, (x \neq -1)

    ដោះស្រាយសមីការ g(x)=0​ យើងទទួលបានរឹសពីរគឺ: x=1,x=-7.

    លើចន្លោះ ( -\infty ,-7), (1, +\infty ) g​ ជាអនុគមន៍ជាប់។

    លើសពីនេះទៀត​​ :

    +g(-8)= \dfrac{-128+11 \sqrt{130}-2}{-16+ \sqrt{130}}= \dfrac{ \sqrt{130}( \sqrt{130}-11)}{16- \sqrt{130}} > 0

    \Rightarrow g(x) > 0, \forall x < -7 \iff f(x) > x, \forall x < -7

    +g(2)= \dfrac{-10+ \sqrt{10}}{ \sqrt{10}+4} < 0 \Rightarrow g(x) < 0, \forall x > 1 \Rightarrow f(x) < x, \forall x > 1.

    **តាង :h(x)=f(x)-1= \dfrac{2 \sqrt{2x^{2}+2}-2(x+1)}{2x+ \sqrt{2x^{2}+2}}, \quad x \neq -1

    សមីការ h(x)=0 មានរឹសមួយគត់គឺ​ x=1.
    នាំអោយ h​ មិនផ្តូរសញាលើចន្លោះ ( -\infty ,1),(1, +\infty ).

    លើសពីនេះទៀត:

    \left\{ \begin{array}{l} h(0)= \dfrac{2 \sqrt{2}--2}{ \sqrt{2}}=2- \sqrt{2} > 0 \\ h(2)= \dfrac{2 \sqrt{10}-6}{4+ \sqrt{10}}= \dfrac{ \sqrt{10}-3}{2+ \sqrt{10}} > 0 \end{array} \right. \Rightarrow h(x) \geq 0, \forall x > 1.

    សញ្ញា “=” \iff x=1 \Rightarrow f(x) \geq 1 , \forall x > -1

    ***តាង k(x)=f(x)-(-7)= \dfrac{14x-2+10 \sqrt{2x^{2}+2}}{2x+ \sqrt{2x^{2}+2}}, x \neq -1

    សមីការ k(x)=0 មានរឹសមួយគត់គឺ​ x=-7.
    នាំអោយk(x)មិនដូរសញាលើចន្លោះ ( -\infty,-7),(-7, +\infty ).

    លើសពីនេះទៀត:

    \left\{ \begin{array}{l} k(-8)= \dfrac{10 \sqrt{130}-114}{ \sqrt{130}-16} < 0 \\ k(-6)= \dfrac{10 \sqrt{74}-86}{ \sqrt{74}-12} < 0 \end{array} \right. \Rightarrow k(x) \leq 0, \forall x < -1

    សញ្ញា “=” \iff x=-7 \Rightarrow f(x) \leq -7, \forall x < -1

    តាមបណ្តាលទ្ធផលខាងលើយើងទទួលបាន:

    a) បើ x_{1}=a > -1នោះតាម**​ យើងមាន: x_{2}=f(x_{1}) \geq 1

    ពីនោះនាំអោយ :x_{n} \geq 1, \forall n \geq 2 \quad (''='' \iff a=1 )

    រួមជាមួយ *​ យើងបាន:x_{n+1}=f(x_{n}) \leq x_{n}, \forall n \geq 2 \quad ''='' \iff a=1).

    ដូចនេះស្វ៊ីត{x_{n}}ជាស្វ៊ីតចុះ(បើ a=1នោះ {x_{n}}ជាស្វ៊ីតថេរ) និងទាល់ក្រោម។ ហេតុនោះស្វ៊ីត {x_{n}}ជាស្វ៊ីតបង្រួមហើយ \displaystyle \lim_{n \to +\infty}x_{n}=1

    b) បើx_{1}=a < -1,​ ស្រដៀងករណីខាងលើ​តាម*** យើងបាន:x_{2}=f(x_{1}) \leq-7.

    ពីនោះ \Rightarrow x_{n} \leq -7, \forall n \in N \quad (''='' \iff a=-7)

    រួមជាមួយ * យើងបាន: x_{n+1}=f(x_{n}) \geq x_{n}, \forall x \geq 2 \quad (''='' \iff a=-7)

    ដូចនេះស្វ៊ីត {x_{n}}ជាស្វ៊ីតកើន(បើa=-7 នោះ {x_{n}}ជាស្វ៊ីតថេរ) និងទាល់លើ។
    ហេតុនោះស្វ៊ីត {x_{n}}បង្រួមហើយ \displaystyle \lim_{n \to +\infty}x_{n}=-7.

    13- គេអោយស្វ៊ីត {u_{n}}កំនត់ដូចខាងក្រោម:

    u_{1}=1; \quad u_{n+1}=u_{n}+ \dfrac{u_{n}^{2}}{1999}

    គណនា:\displaystyle \lim_{n \to +\infty} \biggl ( \dfrac{u_{1}}{u_{2}} + \dfrac{u_{2}}{u_{3}}+...+ \dfrac{u_{n}}{u_{n+1}} \biggr ).

  • សំរាយ
  • យើងមាន:\dfrac{u_{n}}{u_{+1}}= \dfrac{u_{n}^{2}}{u_{n+1}u_{n}}= \dfrac{1999(u_{n+1}-u_{n})}{u_{n+1}u_{n}}=1999 \biggl ( \dfrac{1}{u_{n}}- \dfrac{1}{u_{n+1}} \biggr )

    \Rightarrow \dfrac{u_{1}}{u_{2}}+ \dfrac{u_{2}}{u_{3}}+...+ \dfrac{u_{k}}{u_{k+1}}=1999 \biggl ( \dfrac{1}{u_{1}}- \dfrac{1}{u_{k+1}} \biggr )=1999 \biggl (1- \dfrac{1}{u_{k+1}} \biggr )

    លើសពីនេះទៀត:u_{n+1} > u_{n} \geq 1, \forall n \in N

    \Rightarrow \begin{Bmatrix} u_{n} \end{Bmatrix}ជាស្វ៊ីតម៉ូណូតូនកើន។
    ហេតុនោះ​,បើស្វ៊ីត{u_{n}}ទាល់លើនោះវាបង្រួមទៅរក​ a កំនត់មួយ:

    a= \displaystyle \lim_{n \to +\infty} u_{n+1}= \lim_{n \to +\infty} (u_{n}+ \dfrac{u_{n}^{2}}{1999})=a+ \dfrac{a^{2}}{1999}

    \Rightarrow a=0

    \Rightarrowមិនពិត (ព្រោះ u_{n} \geq 1, \forall n \in N \Rightarrow a \geq 1)

    ដូចនេះស្វ៊ីត {u_{n}}មិនទាល់លើ.នោះគេបាន:

    \displaystyle \lim_{n \to +\infty} u_{n}= +\infty

    \Rightarrow \displaystyle \lim_{n \to +\infty} \biggl ( \dfrac{u_{1}}{u_{2}}+ \dfrac{u_{2}}{u_{3}}+...+ \dfrac{u_{n}}{u_{n+1}} \biggr )=1999.

    14- គេអោយ p ជាចំនួនបថម.

    បង្ហាញថា \dfrac{C_{2p}^{p}-2}{p^{2}}ជាចំនួនគត់។

  • សំរាយ
  • +បើ p=2 នោះ \dfrac{C_{4}^{2}-2}{4}=1

    +បើ p>2 នោះ: C_{2p}^{p}= \dfrac{(2p)!}{(p!)^{2}}= \dfrac{(2p)(2p-1)!}{p.(p-1)!p!}=2.C_{2p-1}^{p-1}

    លើសពីនេះទៀត:

    (2p-k)(p+k) \equiv k(p-k) (mod \quad p^{2}), (\forall k =1,2,..., \dfrac{p-1}{2})

    \Rightarrow [(2p-1)(p+1).[(2p-2)(p+2)]...[(2p- \dfrac{p-1}{2})(p+ \dfrac{p-1}{2})]

    \equiv [1(p-1)].[2(p-2)]...[ \dfrac{p-1}{2}. \dfrac{p+1}{2}] (mod \quad p^{2})

    \Rightarrow (p+1)(p+2)...(2p-1) \equiv (p-1)! (mod \quad p^{2}  )

    \Rightarrow C_{2p-1}^{p-1}= \dfrac{(p+1)(p+2)...(2p-2)(2p-1)}{(p-1)!}= \dfrac{m.p^{2}+(p-1)!}{(p-1)!},(m \in Z)

    = \dfrac{mp^{2}}{(p-1)!}+1

    \Rightarrow \dfrac{mp^{2}}{(p-1)!}=C_{2p-1}^{p-1}-1គឺជាចំនួនគត់

    m \vdots (p-1)!​ (ព្រោះp^{2},(p-1)!ជាចំនួនបថមរួមគ្នា)

    \Rightarrow m=n.(p-1)! ,(n \in Z)

    \Rightarrow C_{2p-1}^{p-1}-1=n p^{2}

    \Rightarrow C_{2p}{p}-2=2(C_{2p-1}{p-1}-1)=2np^{2}

    \Rightarrow \dfrac{C_{2p}{p}-2}{p^{2}}=2n \in Z.

    15- គេអោយស្វ៊ីតចំនួនពិតវិជ្ជមាន u_{o},u_{1},...,u_{1999} ផ្ទៀងផ្ទាត់លក្ខខ័ណ្ឌ:

    \left\{ \begin{array}{l} u_{o}=u_{1999}=1 \quad (1) \\ u_{1}=2 \sqrt[4]{u_{i-1}.u_{i+1}};i=1,2,...,1998 \quad (2) \end{array} \right.

    បង្ហាញថា :

    a)1 \leq u_{1} < 4; \forall i =0,2,...,1999

    b)u_{o}=u_{1999},u_{1}=u_{1998},u_{999}=u_{1000}

    c)u_{o} < u_{1} < ... < u_{999}.

  • សំរាយ
  • a)តាង \alpha = Max u_{i}, \beta =Minu_{i};i= \overline{0,1999},( \alpha , \beta > 0)

    បើ \alpha = \betaនោះ u_{o}=u_{1}=...=u_{1999}=1;​ តាមប៉ុន្តែលក្ខខ័ណ្ឌ (2) យើងឃើញថាករណីនេះមិនពិត។
    ដូចនេះ :\alpha > \beta .

    ក្រៅពីនេះតាមវិយមន័យគេបាន:

    \alpha \geq u_{i} \geq \beta, \forall i = \overline{1,1999}

    \Rightarrow \alpha \geq 1 \geq \beta \quad (3)

    បើ \beta =u_{k},(k \in \begin{Bmatrix} 1,2,...,1998 \end{Bmatrix} )​ តាម (2) នាំអោយ:

    \beta =u_{k}=2 \sqrt[4]{u_{k-1}u_{k+1}} \geq 2 \sqrt[4]{ \beta^{2}}

    \Rightarrow \beta^{4} \geq 16 \beta^{2} \Rightarrow \beta \geq 4

    \Rightarrow​ មិនពិត (ព្រោះផ្ទុយជាមួយ​(3))

    ដូចនេះ​ \beta \neq u_{k}, \forall k = \overline{0,1998}

    នាំអោយ: \beta =u_{1}=u_{1999}=1

    ដូចនេះយើងបានស្រាយឃើញថា u_{i} \geq 1, \forall i = \overline{0,1999}

    ដោយ: \alpha > \beta =1​ ទើបយើងមាន \alpha =u_{k}ណាមួយ(k \in \begin{Bmatrix} 1,2,...,1998 \end{Bmatrix})

    \Rightarrow \alpha =u_{k}= 2 \sqrt[4]{u_{k-1}.u_{k+1}} \leq 2 \sqrt[4]{ \alpha^{2}}

    \Rightarrow \alpha \leq 4

    បើ​ \alpha =4 \Rightarrow u_{k-1}=u_{k+1}=4.​ ដូចគ្នានឹងស្វ៊ីតជាប់បន្តបន្ទាប់ (2) គេបាន:

    u_{o}=u_{1}=u_{2}=...=u_{1999}=4

    លក្ខខ័ណ្ឌនេះផ្ទុយពីសម្មតិកម្ម (1),នោះ​ \alpha < 4.

    មានន័យថា​ u_{i} < 4, \forall i = \overline{0,1999}.

    សរុបមកយើងសន្និដ្ឋានបានថា:

    1 \leq u_{i} < 4, \forall i = \overline{0,1999}.

    b)ពិនិត្យស្វ៊ីត u_{1999},u_{1998},...u_{1},u_{o}

    ស្វ៊ីតនេះ ក៏ផ្ទៀងផ្ទាត់លក្ខខ័ណ្ឌ (1),(2)។ដោយសារតែលក្ខណះមានតែមួយគត់គេបាន:

    u_{o}=u_{1999}, u_{1}=u_{1998},...,u_{999}=u_{1000}.

    c)តាមលក្ខខ័ណ្ឌ(2)យើងបាន: u_{999}= 2 \sqrt[4]{u_{998}u_{1000}}

    តាមការស្រាយខាងលើយើងមាន:0 < u_{999}=u_{1000} < 4

    \Rightarrow u_{999}^{2}=4 \sqrt{u_{998}.u_{999}} < 4u_{999}

    \Rightarrow u_{998}.u_{999} < u_{999}^{2} \Rightarrow u_{998} < u_{999}

    ធ្វើការសន្និដ្ឋានដូចគ្នាយើងនឹងទទួលបានលក្ខខ័ណ្ឌដែលត្រូវស្រាយ។

    16- ពីស្វ៊ីត {u_{n}}កំនត់ដោយ:

    \left\{ \begin{array}{l} u_{1}=2 \\ u_{n+1}= \dfrac{u_{n}^{2}+1999u_{n}}{2000} , n \in N \end{array} \right.

    យើងបង្កើតបានស្វ៊ីត​ {S_{n}} ជាមួយ S_{n}= \displaystyle \sum_{i=1}^{n} \dfrac{u_{i}}{u_{i+1}-1}.

    គណនា:\displaystyle \lim_{n \to +\infty} S_{n}.

  • សំរាយ
  • តាមសម្មតិកម្មគេបាន:

    u_{n+1}= \dfrac{u_{n}(u_{n}-1)}{2000}+u_{n}, \forall n \in N.

    ដោយ u_{1}=2នោះយើងមាន:2=u_{1} < u_{2} < ... < u_{n} < ...

    មានន័យថា{u_{n}}ជាស្វ៊ីតមួយកើន។

    ឧបមាថាស្វ៊ីតនេះទាល់លើ នោះមាន L \in [2, +\infty ) : \displaystyle \lim_{n \to +\infty} u_{n}=L

    នោះគេទាញបាន:L= \dfrac{L^{2}+1999L}{2000} \iff \left\{ \begin{array}{l} L=0 \\ L=1 \end{array} \right. មិនពិត(ព្រោះ L \geq 2).

    ដូចនោះស្វ៊ីត {u_{n}}មិនទាល់លើ។

    ហេតុនោះ:\displaystyle \lim_{n \to +\infty} u_{n}= +\infty

    ក្រៅពីនេះតាមសម្មតិកម្មដែរគេបាន:

    u_{n+1}= \dfrac{u_{n}^{2}+1999u_{n}}{2000}

    \Rightarrow u_{n}(u_{n}-1)=2000(u_{n+1}-u_{n})

    \Rightarrow \dfrac{u_{n}}{u_{n+1}-1}= \dfrac{u_{n}(u_{n}-1)}{(u_{n+1}-1)(u_{n}-1)}= \dfrac{2000(u_{n+1}-u_{n})}{(u_{n+1}-1)(u_{n}-1)}=2000( \dfrac{1}{u_{n}-1}- \dfrac{1}{u_{n+1}-1})

    \Rightarrow S_{n}= \displaystyle \sum_{i=1}^{n} \dfrac{u_{i}}{u_{n+1}-1}=2000( \dfrac{1}{u_{1}-1}- \dfrac{1}{u_{n+1}-1})=2000(1- \dfrac{1}{u_{n+1}-1})

    \Rightarrow \displaystyle \lim_{n \to +\infty} S_{n}=2000.

    17- គេអោយស្វ៊ីតចំនួនពិត​ {a_{n}}ជាមួយ :\left\{ \begin{array}{l} a_{o}=1 \\ a_{1}=2 \\ a_{n+2}=a_{n+1}-a_{n}, \forall n \in N \end{array} \right.

    រកតំលៃទាំងអស់របស់ n​ ដើម្បីអោយ a_{n}-1ជាចំនួនការេមួយប្រាកដ។

  • សំរាយ
  • ពិនិត្យសមីការតំណាង:

    t^{2}=4t-1 \iff t=2 \pm \sqrt{3}

    \Rightarrow a_{n}= \alpha (2+ \sqrt{3})^{n}+ \beta (2- \sqrt{3})^{n}, ( \alpha, \beta \in R)

    លើសពីនេះទៀត: \left\{ \begin{array}{l} a_{o}=1 \\ a_{1}=2 \end{array} \right.

    \iff \left\{ \begin{array}{l} \alpha + \beta =1 \\ 2( \alpha + \beta )+ \sqrt{3}( \alpha - \beta )=2 \end{array} \right. \iff \alpha = \beta = \dfrac{1}{2} \Rightarrow a_{n}= \dfrac{1}{2}(t_{1}^{n}+t_{2}^{n})

    ដោយ:2 \pm \sqrt{3}= \dfrac{1}{2}( \sqrt{3} \pm 1)^{2}=\biggl ( \dfrac{ \sqrt{3} \pm 1}{ \sqrt{2}} \biggr )^{2}

    \Rightarrow a_{n}-1= \dfrac{1}{2} \biggl[ \biggl ( \dfrac{ \sqrt{3}+1}{ \sqrt{2}}\biggl )^{2n}+ \biggl ( \dfrac{ \sqrt{3}-1}{ \sqrt{2}} \biggr )^{2n} \biggr ]-1= \biggl [ \dfrac{( \sqrt{3}+1)^{n}.( \sqrt{3}-1)^{n}}{( \sqrt{2})^{n+1}} \biggr ]^{2}

    ដោយ a_{n}-1 គឺជាចំនួនការេប្រកដគេបាន:

    A= \dfrac{( \sqrt{3}+1)^{n}-( \sqrt{3}-1)^{n}}{( \sqrt{2})^{n+1}} \in Z

    យើងពិនិត្យករណីរៀងគ្នា:

    +n=0 \Rightarrow A=0 \in Z

    +n=1 \Rightarrow A=1 \in Z

    +n=2k, k \in N^{*}

    ពិនិត្យស្វ៊ីត {b_{k}},ជាមួយ:

    b_{k}= \dfrac{(2+ \sqrt{3})^{k}-(2- \sqrt{3})^{k}}{ \sqrt{2}}= \dfrac{( \sqrt{3}+1)^{n}-( \sqrt{3}-1)^{n}}{(\sqrt{2})^{n+1}}=A

    នោះយើងទទួលបាន 2 \pm \sqrt{3}ជារឹសរបស់សមីការ x^{2}=4x-1
    នោះគេបាន{b_{k}}​​​ ផ្ទៀងផ្ទាត់:b_{k+2}=4b_{k+1}-b_{k}

    ដែល:\left\{ \begin{array}{l} b_{o}=0 \\ b_{1}= \sqrt{6} \end{array} \right. \Rightarrow b_{k} \notin Q, \forall k \in N^{*}

    \Rightarrow a_{n}-1មិនមែនជាចំនួនការេប្រកដទេ។

    +បើ n=2k+1, k \in N^{*}:​ យើងមាន:

    \dfrac{( \sqrt{3}+1)^{n}-( \sqrt{3}-1)^{n}}{( \sqrt{2})^{n+1}}= \dfrac{ \sqrt{3}+1}{2} \biggl [ \biggl ( \dfrac{ \sqrt{3}+1}{ \sqrt{2}} \biggr )^{2k}- \biggl ( \dfrac{ \sqrt{3}-1}{ \sqrt{2}} \biggr )^{2k} \biggr ]

    = \dfrac{ \sqrt{3}+1}{2}[(2+ \sqrt{3})^{k}-(2- \sqrt{3})^{k}]

    តាង:C_{k}= \dfrac{ \sqrt{3}+1}{2}[(2+ \sqrt{3})^{k}-(2- \sqrt{3})^{k}], k \in N^{*}

    នោះស្វ៊ីត {C_{k}}ផ្ទៀងផ្ទាត់ C_{k+1}=4C_{k+1}-C_{k}

    ដែល: \left\{ \begin{array}{l} C_{o}=0 \\ C_{1}=5 \end{array} \right. \Rightarrow C_{k} \in Z, \forall k \in N^{*}

    ដូចនេះ:a_{n}-1គឺជាចំនួនការេប្រកដ \iff n=0 រឺ​ n គត់វិជ្ជមានសេស។

    18- គេអោយ \alpha \in (0, \dfrac{ \pi }{2}). គណនា: \displaystyle \lim_{n \to +\infty} (cos^{2} \alpha \sqrt[n]{cos \alpha }+sin^{2} \alpha \sqrt[n]{sin \alpha })^{n}.

  • សំរាយ
  • តាង: x_{n}=cos^{2} \alpha \sqrt[n]{cos \alpha }+sin^{2} \alpha \sqrt[n]{sin \alpha }, n \in N

    \Rightarrow x_{n} \rightarrow 1,ពេល n \rightarrow +\infty

    \Rightarrow \dfrac{lnx_{n}}{x_{n}-1} \rightarrow n,ពេល n \rightarrow +\infty

    \biggl ( \left\{ \begin{array}{l} o < x_{n} < 1, \forall n \in N \\ \dfrac{ln(1+x)}{x} \rightarrow 1, (x \rightarrow o) \end{array} \right. \biggr )

    \Rightarrow \dfrac{nlnx_{n}}{n(x_{n}-1)} \rightarrow 1, (n \rightarrow +\infty

    ដែល n(x_{n}-1)=cos^{2} \alpha \dfrac{ \sqrt[n]{cos \alpha}-1}{ \dfrac{1}{n}}+sin^{2} \alpha \dfrac{ \sqrt[n]{sin \alpha }-1}{ \dfrac{1}{n}}

    \rightarrow cos^{2} \alpha lncos \alpha +sin^{2} \alpha lnsin \alpha​ (ព្រោះ \displaystyle \lim_{n \to +\infty} n( \sqrt[n]{x}-1)=lnx, (x > 0))

    \Rightarrow (x_{n})^{n} \rightarrow (cos \alpha )^{cos^{2} \alpha}.(sin \alpha)^{sin^{2} \alpha}.

    19- គេអោយស្វ៊ីតចំនួនពិត {u_{n}}ជាមួយ: \left\{ \begin{array}{l} u_{1} \in R \\ u_{n+1}= \dfrac{1}{2}ln(1+u_{n}^{2})-1999, n \geq 1 \end{array} \right.

    បង្ហាញថាស្វ៊ីត {u_{n}}ជាស្វ៊ីតបង្រួម។

  • សំរាយ
  • យើងមាន:f(x)= \dfrac{1}{2}ln(1+x^{2})-1999 ជាអនុគមន៍មានដេរីវេលើ R ហើយ:

    f'(x)= \dfrac{x}{1+x^{2}} \in [- \dfrac{1}{2}, \dfrac{1}{2}] ( \forall x \in R)

    ក្រៅពីនេះយើងតាង: g(x)=x+1999- \dfrac{1}{2}ln(1+x^{2})=x-f(x)

    នោះ g​ ក៏មានដេរីវេលើR​ដែរ ហើយ: g'(x)=1- \dfrac{x}{1+x^{2}} >0, ( \forall x \in R)

    លើសពីនេះទៀត:

    g(0).g(-1999)=- \dfrac{1999}{2}ln(1+1999^{2}) < 0

    ពីនោះនាំអោយមាន L \in (-1999,0)ដែល: g(L)=0 \iff f(L)=L

    អនុវត្តន៍ទ្រឹស្តី Lagrange យើងមាន C \in E ដែល:

    |u_{n+1}-L|=|f(n_{1})-f(L)|=|f'(C)|.|u_{n}-L| \leq \dfrac{1}{2}|u_{n}-L|

    នោះយើងបាន:

    |u_{n}-L| \leq ( \dfrac{1}{2})^{n-1}. |u_{1}-L|, \quad \forall n \in N

    \Rightarrow \displaystyle \lim_{n \to +\infty} u_{n}=L.

    20- គេអោយ \left\{ \begin{array}{l} n \in N \\ n \geq 3 \end{array} \right.

    ចូរកំនត់តំលៃធំបំផុតរបស់G និងតូចបំផុតរបស់ K ដែលជាមួយ ចំនួន n វិជ្ជមានណាមួយនៃ a_{1},a_{2},...a_{n}.​ នោះវិសមភាពខាងក្រោមពិតជានិច្ច:

    K < \dfrac{a_{1}}{a_{1}+a_{2}}+ \dfrac{a_{2}}{a_{2}+a_{3}}+...+\dfrac{a_{n}}{a_{n}+a_{n+1}} < G.

  • សំរាយ
  • តាង: S= \displaystyle \sum_{i=1}^{n}a_{i} និង T=T(a_{1},a_{2},...,a_{n})= \displaystyle \sum_{i=1}^{n} \dfrac{a_{i}}{a_{i}+a_{i+1}}

    (ក្នុងនោះ: a_{n+1}=a_{1})

    យើងមាន: T > \displaystyle \sum_{i=1}^{n} \dfrac{a_{i}}{S}=1 \Rightarrow K \geq 1.

    ក្រៅពីនេះ:

    n-T= \dfrac{a_{2}}{a_{1}+a_{2}}+ \dfrac{a_{2}}{a_{2}+a_{3}}+...+ \dfrac{a_{n}}{a_{n}+a_{n+1}}+ \dfrac{a_{1}}{a_{n}+a_{1}}

    =T(a_{n},a_{n-1},...,a_{1}) \geq 1

    \Rightarrow T < n-1, \quad G \leq n -1

    ជាមួយ x>0,យើងមាន:

    T(1,x,...,x^{n-1})= \dfrac{1}{1+x}+ \dfrac{x}{x+x^{2}}+...+ \dfrac{x^{n-1}}{x^{n-1}+1}= \dfrac{n-1}{1+x}+ \dfrac{x^{n-1}}{1+x^{n-1}}

    \left\{ \begin{array}{l} \displaystyle \lim_{x \to 0^{+}} T(1,x,....x^{n-1})=n-1 \\ \displaystyle \lim_{n \to +\infty} T(1,x,...,x^{n-1})=1 \end{array} \right.

    ដូចនេះ: MaxG=n-1, MinK=1.

    21- គេអោយស៊្វីត {u_{n}}កំនត់ដូចខាងក្រោម:

    \left\{ \begin{array}{l} u_{o}=2000 \\ u_{n+1}=u_{n}+ \dfrac{1}{u_{n}^{2}}; n=0,1,2,... \end{array} \right. គណនាលីមីត: \displaystyle \lim_{n \to +\infty} \dfrac{u_{n}^{3}}{n}.

  • សំរាយ
  • យើងមាន: u_{n+1}=u_{n}+ \dfrac{1}{u_{n}^{2}}, \forall n \geq 0

    \Rightarrow u_{n+1}^{3}=u_{n}^{3}+ \dfrac{1}{u_{n}^{6}}+3+ \dfrac{3}{u_{n}^{3}} > u_{n}^{3}+3, \forall n \geq 0 \quad (1) (ព្រោះ u_{n} > 0, \forall n \geq 0)

    \Rightarrow \left\{ \begin{array}{l} u_{1}^{3} > u_{o}^{3}+3 \\ u_{2}^{3} > u_{1}^{3}+3 \\ ............ \\ u_{n}^{3} > u_{n-1}^{3}+3 \end{array} \right. , \forall n \geq 1

    \Rightarrow u_{n}^{3} > 3n+u_{o}^{3}, \forall n \geq 1 \quad (2)

    តាម(1)​និង(2) គេបាន:

    u_{n+1}^{3} < u_{n}^{3}+3+ \dfrac{1}{u_{o}^{3}+3n}+ \dfrac{1}{(u_{o}^{3}+3n)^{2}} < u_{n}^{3}+3+ \dfrac{1}{n}+ \dfrac{1}{9n^{2}}, \forall n \geq 1

    \Rightarrow u_{n}^{3} < u_{1}^{3} +3(n-1)+ \displaystyle \sum_{k=1}^{n-1} \dfrac{1}{k}+ \dfrac{1}{9} \sum_{k=1}^{n-1} \dfrac{1}{k^{2}}, \forall n \geq 2 ,(3)

    ក្រៅពីនេះយើងមាន:

    \displaystyle \sum_{k=1}^{n} \frac{1}{k^{2}} < 1+ \dfrac{1}{1.2}+ \dfrac{1}{2.3} +...+ \dfrac{1}{n(n-1)}

    \leq 1+(1- \dfrac{1}{2})+( \dfrac{1}{2}- \dfrac{1}{3})+...+( \dfrac{1}{n-1}- \dfrac{1}{n}) < 2, \forall n \geq 1

    \Rightarrow ( \displaystyle \sum_{k=1}^{n} \dfrac{1}{k})^{2} \leq n \sum_{k=1}^{n} \dfrac{1}{k^{2}} < 2n, \forall n \geq 1 ,(4)

    \Rightarrow \displaystyle \sum_{k=1}^{n} \dfrac{1}{k} < \sqrt{2n}, \forall n \geq 1,(5)

    តាម(2),(3),(4) និង (5)គេបាន:

    3+ \dfrac{u_{o}^{3}}{n} < \dfrac{u_{n}^{3}}{n} < \dfrac{u_{1}^{3}}{n}+3 \sqrt{ \dfrac{2}{n}}+ \dfrac{2}{9n}, \forall n \geq 2

    ដោយ :\displaystyle \lim_{n \to +\infty} (3+ \dfrac{u_{o}^{3}}{n})= \lim_{n \to +\infty} ( \frac{u_{1}^{3}}{n}+3+ \sqrt{ \dfrac{2}{n}}+ \frac{2}{9n})=3

    ដូចនេះ: \displaystyle \lim_{n \to +\infty} \dfrac{u_{n}^{3}}{n}=3.

    22- គេអោយស្វ៊ីត​ {u_{n}​}កំនត់ដោយ:

    \left\{ \begin{array}{l} u_{1}=1 \\ u_{2}=2 \\ u_{n+2}=u_{n}+2u_{n+1}, n \in N^{*} \end{array} \right.

    គណនា a ដែល:a= \displaystyle \lim_{n \to +\infty} \dfrac{u_{n+1}}{u_{n}}.

  • សំរាយ
  • +ជាមួយ n=1,យើងមាន:

    u_{2}^{2}-u_{1}u_{3}=4-1.5=(-1)^{1}

    +n=k \in N,ឧបមាថា: u_{k+1}^{2}-u_{k}.u_{k+2}=(-1)^{k}.

    +n=k+1 យើងមាន:

    u_{k+2}^{2}-u_{k+1}.u_{k+3}= u_{k+2}(u_{k}+2u_{k+1})-u_{k+1}(u_{k+1}+2u_{k+2})

    =u_{k+2}.u_{k}-u_{k+1}^{2}=-(-1)^{k}=(-1)^{k+1}

    នោះ:u_{n+1}^{2}-u_{n}.u_{n+2}=(-1)^{n}, \forall n \in N

    ម្យ៉ាងទៀត:u_{n} \geq 1, \forall n \in N,នោះ;

    ( \dfrac{u_{n+1}}{u_{n}})^{2}- \dfrac{u_{n+2}}{u_{n}}= \dfrac{(-1)^{n}}{u_{n}^{2}}

    \Rightarrow ( \dfrac{u_{n+1}}{u_{n}})^{2}- \dfrac{u_{n}+2u_{n+1}}{u_{n}}= \dfrac{(-1)^{n}}{u_{n}^{2}}

    \Rightarrow ( \dfrac{u_{n+1}}{u_{n}})^{2}-2 \dfrac{u_{n+1}}{u_{n}}-1= \dfrac{(-1)^{n}}{u_{n}^{2}}(*)

    លើសពីនេះទៀតពី​:u_{n+2}=u_{n}+2u_{n+1}, \forall n \in N

    \Rightarrow u_{n+2}-u_{n+1}=u_{n}+u_{n+1} > 0, \forall n \in N

    \Rightarrow u_{n+2} > u_{n+1}, \forall n \in N

    \Rightarrow \begin{Bmatrix} u_{n} \end{Bmatrix}ជាស្វ៊ីតកើន.

    ហេតុនោះ បើ{u_{n}}ទាល់នោះ{u_{n}}បង្រួមទៅរក​ b \in R.

    ពេលនោះពី:u_{n+1}=u_{n}+2u_{n+1}, \forall n \in N

    \Rightarrow b=b+2b

    \Rightarrow b=0

    \Rightarrowមិនពិត (ព្រោះ u_{n} \geq 1, \forall n \in N \Rightarrow b \geq 1)

    \Rightarrow \begin{Bmatrix} u_{n} \end{Bmatrix}មិន​ទាល់លើ។

    ដូចនោះ{u_{n}}កើន និងមិនទាល់លើ នោះគេបាន:

    \displaystyle \lim_{n \to +\infty} u_{n}= +\infty.

    រួមជាមួយ(*)នាំអោយ :a_{2}-2a-1=0 \iff a=1+ \sqrt{2},(a \geq 1)

    ដូចនេះ: a=1+ \sqrt{2}.

    23- គេអោយ f:N \rightarrow Z
    . \quad n \rightarrow f(n)ផ្ទៀងផ្ទាត់:

    \left\{ \begin{array}{l} f(m+n)-f(m)-f(n) \in \begin{Bmatrix} 0,1 \end{Bmatrix}, \forall m,n \in N \\ f(n) \geq 0, \forall n \in N \\ f(2)=0 < f(3), f(9999)=3333 \end{array} \right.​គណនា :f(2000)?.

  • សំរាយ
  • ដោយ: f(m+n)-f(m)-f(n) \in \begin{Bmatrix} 0,1 \end{Bmatrix}, \forall m,n \in N

    \Rightarrow f(m+n) \geq f(m)+f(n), \forall m,n \in N

    +យក m=n=1, យើងមាន: 0=f(2) \geq 2f(1) \Rightarrow f(1) \leq 0

    ដែល :f(1) \geq 0(តាមសម្មតិកម្ម) \Rightarrow f(1)=0

    +យក m=2,n=1,យើងមាន:

    f(3)-f(2)-f(1) \in  \begin{Bmatrix} 0,1 \end{Bmatrix}

    \Rightarrow f(3) \in \begin{Bmatrix} 0,1 \end{Bmatrix}(ព្រោះ f(2)=0,f(1)=1)

    តែ f(3)>0 \Rightarrow f(3)=1

    \Rightarrow f(2.3)=f(3+3) \geq f(3)+f(3)=2

    \Rightarrow f(2.3) \geq 2

    ឧបមាថា f(k.3) \geq k (k \in N)

    ពេលនោះ f(k+1).3))=f(k.3+3) \geq f(k.3)+f(3) \geq k+1

    ហេតុនោះយើងមាន:f(3.n) \geq n, \forall n \in N

    ម្យ៉ាងទៀត: បើ \left\{ \begin{array}{l} n \in N \\ f(3n) > n \end{array} \right.

    \Rightarrow f(3(n+1))=f(3n+3) \geq f(3n)+f(3) > n+1

    នោះ​ f(3m) > m, \forall m \geq n

    ប៉ុន្តែដោយ: f(9999)=f(3.3333)=3333នោះ:

    \Rightarrow f(3n)=n, \forall n \in \begin{Bmatrix} 1,2,...,3333 \end{Bmatrix}

    \Rightarrow f(3.2000)=2000=f(3.2000 \geq f(2.2000+f(2000) \geq 3f(2000)

    \Rightarrow f(2000) \leq \dfrac{2000}{3} < 667

    ក្រៅពីនេះ: f(2000) \geq f(1998)+f(2) \geq f(3.666)=666

    \Rightarrow f(2000) \geq 666

    ហេតុនោះ: 666 \leq f(2000) < 667

    \Rightarrow f(2000)=666 (ព្រោះ​f(2000) \in Z.)

    24- គេអោយស្វ៊ីតចំនួនពិត {u_{n}}កំនត់ដោយ:

    \left\{ \begin{array}{l} u_{o} > 0 , u_{1} >0 \\ u_{n+2}=(u_{n+1}.u_{n}^{2})^{ \frac{1}{3}}, \forall n \in N \end{array} \right.

    ចូររកតួទូទៅរបស់ u_{n}.

  • សំរាយ
  • +តាមវិចារកំនើនមានកំនត់យើងមាន:u_{n} > 0, \forall n \in N, u_{n+2}=(u_{n+1}.u_{n}^{2})^{ \frac{1}{3}}

    \Rightarrow lnu_{n+2}= \dfrac{1}{3}lnu_{n+1}+ \dfrac{2}{3}lnu_{n}

    តាង: v_{n}=lnu_{n}, \forall n \geq 0

    \Rightarrow v_{n+2}= \dfrac{1}{3}v_{n+1}+ \dfrac{2}{3}v_{n}

    \iff 3v_{n+2}=v_{n+1}+2v_{n}

    ពិនិត្យសមីការតំណាង:

    3x^{2}-x-2=0 \iff x=1 \bigvee x=- \dfrac{2}{3}

    \Rightarrow v_{n}= \alpha_{1}+ \alpha_{2}(- \dfrac{2}{3})^{n}, ( \alpha_{1}, \alpha_{2} \in R)

    \left\{ \begin{array}{l} \alpha_{1}+ \alpha_{2}=v_{o} \\ \alpha_{1}- \dfrac{2}{3} \alpha_{2}=v_{1} \end{array} \right. \iff \left\{ \begin{array}{l} \alpha_{1}= \dfrac{1}{5}(2v_{o}+3v_{1}) \\ \alpha_{2}= \dfrac{3}{5}(v_{o}-v_{1}) \end{array} \right.

    \Rightarrow v_{n}= \dfrac{1}{5}(2v_{o}+3v_{1})+ \dfrac{1}{5}(3v_{o}-3v_{1})(- \dfrac{2}{3})^{n}

    u_{n}=e^{ \dfrac{1}{5}(2v_{o}+3v_{1})+ \dfrac{1}{5}(3v_{o}-3v_{1})(- \dfrac{2}{3})^{n}}=e^{ \dfrac{1}{5}ln(u_{o}^{2}.u_{1}^{2})+ \dfrac{1}{5}(ln \dfrac{u_{o}^{3}}{u_{1}^{3}})(- \dfrac{2}{3})^{n}}

    u_{n}=(u_{o}^{2}.u_{1}^{3})^{ \dfrac{1}{5}}. u_{o}^{- \dfrac{3}{5}(- \dfrac{2}{3})^{n}}.u_{1}^{- \dfrac{3}{5}(- \dfrac{2}{3})^{n}}=u_{o}^{ \dfrac{2}{5}+ \dfrac{(-2)^{n}}{5.3^{n-1}}}.u_{1}^{ \dfrac{3}{5}- \dfrac{(-2)^{n}}{5.3^{n-1}}}.

    25- គេអោយ a>2. ប្រើវិចារកំនើនមានកំនត់:

    a_{o}=1,a_{1}=a, a_{n+1}= ( \dfrac{a_{n}^{2}}{a_{n-1}^{2}}-2)a_{n}, \forall n \in N.

    បង្ហាញថា:

    \dfrac{1}{a_{o}}+ \dfrac{1}{a_{1}}+...+ \dfrac{1}{a_{n}} < \dfrac{1}{2}(2+a- \sqrt{a^{2}-4}), \forall n \in N.

  • សំរាយ
  • ដោយ a>2, នោះយើងអា​ចរើសបានចំនួយ b > 1 :a=b+ \dfrac{1}{b}

    \Rightarrow a^{2}-2=b^{2}+ \dfrac{1}{b^{2}}

    តាមវិធីកំនត់ស្វ៊ីតយើងមាន:

    a_{2}=( \dfrac{a_{1}^{2}}{a_{o}^{2}}-2)a_{1}=( \dfrac{a^{2}}{1}-2)a=(b^{2}+ \dfrac{1}{b^{2}})(b+ \dfrac{1}{b})

    ឧបមាថា: a_{k}=(b^{2^{k-1}}+ \dfrac{1}{b^{2^{k-1}}})...(b^{2}+ \dfrac{1}{b^{2}})(b+ \dfrac{1}{b}), (k \in N)

    ពេលនោះ:

    a_{k+1}= ( \dfrac{a_{k}^{2}}{a_{k-1}^{2}}-2)a_{k}= \biggl [ \biggl (b^{2^{k-1}}+ \dfrac{1}{b^{2^{k-1}}} \biggr )^{2}-2 \biggr ]a_{k}

    =(b^{2^{k}}+ \dfrac{1}{b^{2^{k}}})...(b^{2}+ \dfrac{1}{b^{2}})(b+ \frac{1}{b})

    ដូចនោះ: a_{n}=(b^{2^{n-1}}+ \dfrac{1}{b^{2^{n-1}}})...(b^{2}+ \dfrac{1}{b^{2}})(b+ \dfrac{1}{b}), \forall n \in N

    \Rightarrow \displaystyle \sum_{k=0}^{n} \dfrac{1}{a_{k}}=1+ \frac{b}{b^{2}+1}+ \frac{b^{3}}{(b^{2}+1)(b^{4}+1)}+...+ \frac{b^{2^{n}-1}}{(b^{2}+1)(b^{4}+1)...(b^{2^{n}}+1)}.

    ម្យ៉ាងទៀត:

    \dfrac{1}{2}(2+a- \sqrt{a^{2}-4}= \dfrac{1}{2}(2+b+ \dfrac{1}{b}- \sqrt{(b+ \dfrac{1}{b})^{2}-4})

    = \dfrac{1}{2}(2+b+ \dfrac{1}{b}-b+ \dfrac{1}{b})=1+ \dfrac{1}{b}

    វិសមភាពដែលត្រូវស្រាយសមមូលជាមួយ:

    \dfrac{b^{2}}{b^{2}+1}+ \dfrac{b^{4}}{(b^{2}+1)(b^{4}+1)}+...+ \dfrac{b^{2^{n}}}{(b^{2}+1)...(b^{2^{n}}+1)} < 1

    \iff 1- \dfrac{1}{(b^{2}+1)(b^{4}+1)...(b^{2^{n}}+1)} < 1(ពិតជានិច្ច)

    26- ស្វ៊ីតចំនួនពិត​ u_{o},u_{1},...,u_{n}កំនត់ដោយ:

    \left\{ \begin{array}{l} u_{o}= \dfrac{1}{2} \\ u_{k}=u_{k-1}+ \dfrac{1}{n}u_{k-1}^{2};k=1,2,...,n \end{array} \right.

    បង្ហាញថា: 1- \dfrac{1}{n} < u_{n} < 1.

  • សំរាយ
  • យើងមាន: u_{k}=u_{k-1}+ \dfrac{1}{n}u_{k-1}^{2}; \quad \biggl ( \left\{ \begin{array}{l} n \in N \\ k=1,2,...,n \end{array} \right. \biggr )

    \iff n(u_{k}-u_{k-1})=u_{k-1}^{2}

    \iff nu_{k}-nu_{k-1}+u_{k}.u_{k-1}-u_{k-1}^{2}=u_{k}.u_{k-1}

    \iff u_{k}(n+u_{k-1})-u_{k-1}(n+u_{k-1})=u_{k}.u_{k-1}

    \iff (u_{k}-u_{k-1})(n+u_{k-1})=u_{k}.u_{k-1}

    \iff \dfrac{1}{u_{k-1}}- \dfrac{1}{u_{k}}= \dfrac{1}{n+u_{k-1}}

    ដោយ: \dfrac{1}{2}=u_{o} < u_{1} < ... < u_{n} \Rightarrow \dfrac{1}{u_{k-1}}- \dfrac{1}{u_{k}} < \dfrac{1}{n}

    \Rightarrow \displaystyle \sum_{k=1}^{n} \biggl ( \frac{1}{u_{k-1}}- \frac{1}{u_{k}} \biggr ) < n. \frac{1}{n} \Rightarrow \dfrac{1}{u_{o}}- \frac{1}{u_{n}} < 1

    \Rightarrow \dfrac{1}{u_{n}} > \dfrac{1}{u_{o}}-1=1 \Rightarrow \dfrac{1}{2} < u_{o} < 1 \quad (1)

    \Rightarrow \dfrac{1}{2} \leq u_{k-1} < 1 \Rightarrow \dfrac{1}{u_{k-1}}- \dfrac{1}{u_{k}}= \dfrac{1}{n+u_{k-1}} > \dfrac{1}{n+1}

    \Rightarrow \displaystyle \sum_{k=1}^{n} \biggl ( \dfrac{1}{u_{k-1}}- \dfrac{1}{u_{k}} \biggr ) > \frac{n}{n+1} \Rightarrow \dfrac{1}{u_{o}}- \dfrac{1}{u_{n}} > \frac{n}{n+1}

    \Rightarrow 2- \dfrac{n}{n+1} > \dfrac{1}{u_{n}} \Rightarrow \dfrac{n+2}{n+1} > \dfrac{1}{u_{n}}

    \Rightarrow u_{n} > \dfrac{n+1}{n+2} > \dfrac{n-1}{n} \Rightarrow u_{n} > 1- \dfrac{1}{n}, \quad (2)

    តាម(1) និង(2): \Rightarrow 1- \dfrac{1}{n} < u_{n} < 1.

    27- គេអោយស៊្វីតចំនួនពិត{u_{n}}ផ្ទៀងផ្ទាត់:

    u_{n}= \sqrt[4]{a_{1}+ \sqrt[4]{a_{2}+...+ \sqrt[4]{a_{n}}}}, 0 < a_{n} < 2000.30^{4^{n}}, n \in N

    ស្រាយបញ្ជាក់ថា{u_{n}}ជាស្វ៊ីតបង្រួម។

  • សំរាយ
  • .+តាង:b_{n}= \dfrac{a_{n}}{30^{4^{n}}}, n \in N \Rightarrow 0 < b_{n} < 2000

    +ពិនិត្យ:\begin{Bmatrix} V_{n}= \sqrt[4]{b_{1}+ \sqrt[4]{b_{2}+...+ \sqrt[4]{b_{n}}}} \end{Bmatrix}

    \Rightarrow u_{n}=30V_{n}; \forall n \in Nនិង {V_{n}}ជាស្វ៊ីតកើន។

    +ពិនិត្យស៊្វីត \begin{Bmatrix} W_{n} \end{Bmatrix}: \left\{ \begin{array}{l} W_{1}= \sqrt[4]{2000} \\ W_{n}= \sqrt[4]{2000+u_{n-1}}, n \geq 2 \end{array} \right.

    *{W_{n}}ជាស្វ៊ីតកើន

    *ពិនិត្យ f(x)=x^{4}-x-2000, x \geq \sqrt[4]{2000}

    f'(x)=4x^{3}-1 > 0, \forall x \geq \sqrt[4]{2000}

    \Rightarrow fកើន លើចន្លោះ [ \sqrt[4]{2000}, +\infty)

    លើសពីនេះទៀត:

    f( \sqrt[4]{2000}).f( \sqrt{2000})=- \sqrt[4]{2000}(2000^{2}-2000- \sqrt{2000}) < 0

    \Rightarrow \exists !c \in ( \sqrt[4]{2000}, \sqrt{2000}): f(c)=0

    \Rightarrow c^{4}=c+2000

    \Rightarrow W_{n} < c, \forall n \in N

    \Rightarrow \begin{Bmatrix} W_{n} \end{Bmatrix}​ បង្រួម

    +W_{n}= \sqrt[4]{2000+ \sqrt[4]{2000+...+ \sqrt[4]{2000}}} > V_{n}, \forall n \in R

    \Rightarrow \begin{Bmatrix} V_{n} \end{Bmatrix}ជាស្វ៊ីតកើនទាល់លើ

    \Rightarrow \begin{Bmatrix} u_{n} \end{Bmatrix}ជាស្វ៊ីតកើនទាល់លើ នោះវាបង្រួម។

    28- ​ គេអោយស្វ៊ីតចំនួនពិត{u_{n}}កំនត់ដោយ:

    \left\{ \begin{array}{l} u_{o}=2 \\ u_{n+1}=3u_{n}+ \sqrt{8u_{n}^{2}+1},n \in N \end{array} \right.

    រករូបមន្តរបស់តួទូទៅនៃ{u_{n}}.

  • សំរាយ​
  • យើងមាន: u_{n+1}=3u_{n}+ \sqrt{8u_{n}^{2}+1}

    \Rightarrow (u_{n+1}-3u_{n})^{2}=8u_{n}^{2}+1

    \Rightarrow u_{n+1}^{2}-6u_{n+1}.u_{n}+9u_{n}^{2}=8u_{n}^{2}+1

    \Rightarrow u_{n+1}^{2}-6u_{n+1}.u_{n}+u_{n}^{2}=1, \quad (1)

    ប្តូរ n+1​ ដោយ n យើងបាន:

    u_{n}^{2} -u_{n}.u_{n-1}+u_{n-1}^{2}=1, \quad (2)

    យក(1) ដក(2) តាមអង្គគេបាន:

    (u_{n+1}^{2}-u_{n-1}^{2})-6u_{n}(u_{n+1}-u_{n-1})=0

    \Rightarrow (u_{n+1}-u_{n-1})(u_{n+1}+u_{n-1}-6u_{n})=0,\quad (3)

    ដោយ: u_{n+1}=3u_{n}+ \sqrt{8u_{n}^{2}+1} > 3u_{n} > 0 , \forall n \geq 0

    \Rightarrow u_{n} > u_{n-1} > 0, \forall n \geq 1

    តាម(3)​ \Rightarrow u_{n+1}-6u_{n}+u_{n-1}=0

    ហេតុនោះស្វ៊ីត {u_{n}}ត្រូវបានកំនត់ដូចខាងក្រោម:

    \left\{ \begin{array}{l} u_{o}=2, \quad u_{1}=6+ \sqrt{33} \\ u_{n+1}=6u_{n}-u_{n-1}, n \geq 1 \end{array} \right.

    សមីការតំណាង​គឺ:

    x^{2}-6x+1=0 \iff x=3+ \sqrt{8};x=3- \sqrt{8}

    \Rightarrow u_{n}= \alpha (3+ \sqrt{8})^{n}+ \beta (3- \sqrt{8})^{n}

    \Rightarrow \left\{ \begin{array}{l} u_{o}= \alpha + \beta =2 \\ u_{1}= \alpha (3+ \sqrt{8})+ \beta (3- \sqrt{8})=6+ \sqrt{33} \end{array} \right.

    \Rightarrow \left\{ \begin{array}{l} \alpha = \dfrac{8+ \sqrt{66}}{8} \\ \beta = \dfrac{8- \sqrt{66}}{8} \end{array} \right.

    ដូចនេះ: u_{n}= \dfrac{8+ \sqrt{66}}{8}(3+ \sqrt{8})^{n}+ \dfrac{8- \sqrt{66}}{8}(3- \sqrt{8})^{n}.

    29- គេអោយស្វ៊ីតចំនួនពិត{x_{n}}កំនត់ដោយ:

    x_{n}= \biggl (1+ \dfrac{1}{n^{2}} \biggr ) \biggl (1+ \dfrac{2}{n^{2}} \biggr )... \biggl (1+ \dfrac{n}{n^{2}} \biggr )

    រក \displaystyle \lim_{n \to +\infty} (lnx_{n}).

  • សំរាយ
  • ដំបូងយើងបង្ហាញវីសមភាពខាងក្រោមសិន:

    x- \dfrac{x^{2}}{2} < ln(1+x) < x, \forall x > 0

    ពិនិត្យ: \left\{ \begin{array}{l} f(x)=ln(1+x)-x+ \dfrac{x^{2}}{2} \\ g(x)=x-ln(1+x) \end{array} \right. (x > 0)

    \Rightarrow \left\{ \begin{array}{l} f'(x)= \dfrac{1}{1+x}-1+x= \dfrac{x^{2}}{x+1} > 0 \\ g'(x)=1- \dfrac{1}{1+x}= \dfrac{x}{x+1} > 0 \end{array} \right. ( \forall x > 0)

    \Rightarrow f,g សុទ្ធតែកើនលើចន្លោះ (0, +\infty )

    \left\{ \begin{array}{l} f(x) > f(0)=0 \\ g(x) > g(0)=0 \end{array} \right. ( \forall x > 0)

    នោះគេបាន: x- \dfrac{x^{2}}{2} < x,  \forall x > 0

    រក \displaystyle \lim_{n \to +\infty} (lnx_{n}):

    យើងមាន: lnx_{n}=ln(1+ \dfrac{1}{n^{2}})+ ln (1+ \dfrac{2}{n^{2}})+...+ln(1+ \dfrac{n}{n^{2}}

    អនុវត្តន៍វិសមភាពខាងលើជាមួយ: \left\{ \begin{array}{l} x= \dfrac{1}{n^{2}} \\ i=1,2,...,n \end{array} \right.

    យើងមាន: \dfrac{i}{n^{2}}- \dfrac{i^{2}}{2n^{4}} < ln(1+ \dfrac{i}{n^{2}}) < \dfrac{i}{n^{2}}

    \Rightarrow \dfrac{1}{n^{2}}(1+2+...+n)- \dfrac{1}{2n^{4}}(1^{2}+2^{2}+...+n^{2}) < lnx_{n} < \dfrac{1}{n^{2}}(1+2+...+n)

    \iff \dfrac{1}{n^{2}}. \dfrac{n(n+1)}{2}- \dfrac{1}{2n^{4}}. \dfrac{n(n+1)(2n+1)}{6} < lnx_{n} < \dfrac{1}{n^{2}}. \dfrac{n(n+1)}{2}

    តែ: \displaystyle \lim_{n \to +\infty} \dfrac{1}{n^{2}}. \dfrac{n(n+1)}{2}- \dfrac{1}{2n^{4}}. \dfrac{n(n+1)(2n+1)}{6}= \lim_{n \to +\infty} \dfrac{1}{n^{2}}. \dfrac{n(n+1)}{2} =  \dfrac{1}{2}

    \Rightarrow \displaystyle \lim_{n \to +\infty} ln(x_{n})= \dfrac{1}{2}.

    30- គេអោយស្វ៊ីត{u_{n}}កំនត់ដោយ:

    \left\{ \begin{array}{l} u_{1}=2 \\ u_{2}=8 \\ u_{n}=4u_{n-1}-u_{n-2} > 0, n \geq 3 \end{array} \right. និង S_{n}= \displaystyle \sum_{i=1}^{n}arccotg(u_{i}^{2}).

    រក \displaystyle \lim_{n \to +\infty} S_{n}.

  • សំរាយ
  • ដំបូងយើងបង្ហាញ u_{n}^{2}-u_{n+1}.u_{n-1}=4, \forall n \geq 2

    យើងមាន:
    u_{n}(4u_{n-1})=u_{n-1}4u_{n}

    \Rightarrow u_{n}(u_{n}+u_{n-2})=u_{n-1}(u_{n+1}+u_{n-1})

    \Rightarrow u_{n}^{2}-u_{n+1}.u_{n-1}=u_{n-1}^{2}-u_{n}.u_{n-2}

    \Rightarrow u_{n}^{2}-u_{n+1}.u_{n-1}=u_{n-1}^{2}-u_{n}.u_{n-2}=...=u_{2}^{2}-u_{3}.u_{1}=8^{2}-(4.8-2)2=4

    \Rightarrow arcctg u_{n}^{2}=arccotg[u_{n}( \dfrac{4u_{n}}{4})]

    = \dfrac{u_{n}(u_{n+1}+u_{n-1})}{u_{n}^{2}-u_{n+1}.u_{n-1}}=arccotg \dfrac{ \dfrac{u_{n+1}}{u_{n}}. \dfrac{u_{n}}{u_{n-1}}+1}{ \dfrac{u_{n}}{u_{n-1}}- \dfrac{u_{n+1}}{u_{n}}}

    =arccotg \dfrac{u_{n+1}}{u_{n}}-arctg \dfrac{u_{n}}{u_{n-1}}

    \Rightarrow S_{n}= \displaystyle \sum_{i=1}^{n}arccotg(u_{i}^{2})=arccotg (u_{1}^{2})+ \sum_{i=2}^{n}arccotg(u_{i}^{2})=arccotg \dfrac{u_{n+1}}{u_{n}}

    លើសពីនេះទៀត: u_{n}=4u_{n-1}-u_{n-2}

    \Rightarrow 1= \dfrac{4u_{n-1}}{u_{n}}- \dfrac{u_{n-2}}{u_{n-1}}. \dfrac{u_{n-1}}{u_{n}}

    \Rightarrow 1=4x-x^{2}, \quad (x = \displaystyle \lim_{n \to +\infty} \dfrac{u_{n-1}}{u_{n}} \leq 1)

    \Rightarrow x^{2}-4x+1=0

    \Rightarrow x=2- \sqrt{3}

    \Rightarrow \displaystyle \lim_{n \to +\infty} \dfrac{u_{n+1}}{u_{n}}=2 + \sqrt{3}

    \Rightarrow \displaystyle \lim_{n \to +\infty} S_{n}= arctg(2+ \sqrt{3})= \dfrac{ \pi }{12}

    \biggl ( \begin{Bmatrix} \dfrac{u_{n-1}}{u_{n}} \end{Bmatrix}មានលីមីតព្រោះ: \left\{ \begin{array}{l} 0 < \dfrac{u_{n-1}}{u_{n}} < 1 \\ \dfrac{u_{n}}{u_{n+1}} > \dfrac{u_{n-1}}{u_{n}} \end{array} \right. \biggr ).

    31- គេអោយស្វ៊ីតចំនួនពិតផ្ទៀងផ្ទាត់ :x_{1},x_{2},...,x_{n}​ និង y_{1},y_{2},...,y_{n}​ ដែល:

    \left\{ \begin{array}{l} y_{i} \in (a,b); i= \overline{1,n} \\ 0 < x_{i} \notin (a,b), i= \overline{1,n} \\ a > 0 \\ x_{1}+x_{2}+...+x_{n}=y_{1}+y_{2}+...+y_{n} \end{array} \right.

    បង្ហាញថា: x_{1}x_{2}...x_{n} \leq y_{1}y_{2}...y_{n}.

  • ​សំរាយ
  • យើងមាន: x_{1}x_{2}...x_{n} \leq y_{1}y_{2}...y_{n}

    \iff n \sqrt[n]{ \displaystyle \prod_{i=1}^{n} \frac{x_{i}}{y_{i}}} \leq n

    តាមវិសមភាព Cauchy:n \sqrt[n]{ \displaystyle \prod_{i=1}^{n} \frac{x_{i}}{y_{i}}} < \displaystyle \sum_{i=1}^{n} \frac{x_{i}}{y_{i}}

    យើងគ្រាន់តែបង្ហាញថា: \displaystyle \sum_{i=1}^{n} \frac{x_{i}}{y_{i}} \leq n \quad (*)

    ហេតុនោះ: (*) \iff \displaystyle \sum_{i=1}^{n} \frac{x_{i}}{y_{i}}-n \leq 0 \quad (**)

    លើសពីនេះទៀតយើងអាចឃើញ:

    0 < x_{1} \leq x_{2} \leq ... \leq x_{k} < y_{1} \leq y_{2} \leq ... \leq y_{k} < x_{k+1} \leq ... \leq x_{n}

    \Rightarrow \displaystyle \sum_{i=1}^{n} \frac{x_{i}-y_{i}}{y_{i}} \leq \sum_{i=1}^{k} \frac{x_{i}-y_{i}}{y_{k}}+ \sum_{j=k+1}^{n} \frac{x_{j}-y_{j}}{y_{k}} \leq \frac{1}{y_{k}} \biggl ( \sum_{i=1}^{n}x_{i}- \sum_{i=1}^{n}y_{i} \biggr )=0

    \Rightarrow (**) (ពិត) \Rightarrow (*)​(ពិត)

    \Rightarrow លំហាត់ត្រូវបានស្រាយបញ្ជាក់។

    32- គេអោយ a,b \in ]0,1[. ពិនិត្យស្វ៊ីត {u_{n}}កំន់តដោយ:

    u_{o}=a, \quad u_{1}=b, \quad u_{n+2}= \dfrac{1}{3}u_{n+1}^{2}+ \dfrac{2}{3} \sqrt{u_{n}}, \forall n \in N.

    បង្ហាញថាស្វ៊ីត{u_{n}}មានលីមីតកំន់តពេល n \rightarrow +\infty .ចូររកលីមីតនោះ។

  • សំរាយ
  • *ពិនិត្យស្វ៊ីត​ {V_{n}}កំនត់ដោយ: \left\{ \begin{array}{l} V_{o}=Min \begin{Bmatrix} a,b \end{Bmatrix} \\ V_{n+1}= \dfrac{1}{3}V_{n}^{2}+ \dfrac{2}{3} \sqrt{V_{n}}, n \geq 0 \end{array} \right.

    តាមវិចារកំនើនមានកំនត់យើងបង្ហាញបាន: 0 < V_{n} < 1 \quad (1)

    ក្រៅពីនេះ:

    V_{n+1}-V_{n}= \dfrac{1}{3}V_{n}^{2}+ \dfrac{2}{3} \sqrt{V_{n}}-V_{n}

    = \dfrac{1}{3} \sqrt{V_{n}}( \sqrt{V_{n}}-1)^{2}( \sqrt{V_{n}}+2) > 0, \forall n \geq 0

    \Rightarrow \begin{Bmatrix} V_{n} \end{Bmatrix} ​កើន (2)

    តាម(1) និង(2) \Rightarrow \begin{Bmatrix} V_{n} \end{Bmatrix}មានលីមីតកំនត់។

    តាង c= \displaystyle \lim_{n \to +\infty}V_{n},យើងមាន:

    \left\{ \begin{array}{l} c= \dfrac{1}{3}c^{2}+ \dfrac{2}{3} \sqrt{c} \\ 0 < V_{1} \leq c \leq 1 \end{array} \right. \iff c=1 \Rightarrow \displaystyle \lim_{n \to +\infty}V_{n}=1.

    * +n=1 \Rightarrow u_{o}=a \in ]0,1[

    + n=k \in N: ឧបមាថា u_{k} \in ]0,1[

    + n=k+1 យើងមាន: u_{k+1}= \dfrac{1}{3}u_{k}^{2}+ \dfrac{2}{3} \sqrt{u_{k-1}} \in ]0,1[.

    \Rightarrow តាមវិចារកំនើនមានកំនត់យើងមាន: u_{k} \in ]0,1[, \forall n \geq 0

    *បង្ហាញថា V_{n} \leq Min \begin{Bmatrix} u_{2n},u_{2n+1} \end{Bmatrix}, \forall  \geq 0

    ហេតុនោះ:

    +n=0: V_{o} \leq Min \begin{Bmatrix} u_{o}, u_{1} \end{Bmatrix} \iff Min \begin{Bmatrix} a,b \end{Bmatrix} \leq Min \begin{Bmatrix} a,b \end{Bmatrix}(ពិត)

    +n=k \in N :ឧបមាថា: V_{k} \leq Min \begin{Bmatrix} u_{2k}, u_{2k+1} \end{Bmatrix}

    +n=k+1:យើងមាន:

    \left\{ \begin{array}{l} V_{k} \leq u_{2k} \\ V_{k} \leq u_{2k+1} \end{array} \right.

    \left\{ \begin{array}{l} u_{2k+2}= \dfrac{1}{3} u_{2k}^{2}+ \dfrac{2}{3} \sqrt{ u_{2k}} \geq \dfrac{1}{3}V_{k}^{2} + \dfrac{2}{3} \sqrt{V_{k}}=V_{k+1} \\ u_{2k+3}= \dfrac{1}{3} u_{2k+2}^{2}+ \dfrac{2}{3} \sqrt{u_{2k+1}} \geq \dfrac{1}{3}V_{k+1}^{2}+ \dfrac{2}{3} \sqrt{V_{k}} \geq \dfrac{1}{3}V_{k}^{2}+ \dfrac{2}{3} \sqrt{V_{k}} =V_{k+1} \end{array} \right.

    \Rightarrow V_{k+1} \leq Min \begin{Bmatrix} u_{2k+2},u_{2k+3} \end{Bmatrix}

    នោះតាមវិចារកំនើនមានកំនត់យើងមាន:

    V_{n} \leq Min \begin{Bmatrix} u_{2n},u_{2n+1} \end{Bmatrix}, \forall n \geq 0

    តាមបណ្តាលទ្ធផលខាងលើគេបាន:

    \left\{ \begin{array}{l} V_{n} \leq u_{2n} < 1 \\ V_{n} \leq u_{2n+1} < 1 \quad ( \forall n \geq 0) \\ \displaystyle \lim_{n \to +\infty} V_{n}=1 \end{array} \right.

    នោះ \displaystyle \lim_{n \to +\infty}u_{2n}= \lim_{n \to +\infty} u_{2n+1}=1

    \iff \displaystyle \lim_{n \to +\infty} u_{n}=1.

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