កំណែ​លំហាត់​សិស្ស​ពូកែ​ភ្នំពេញ ២០១១ (១២) ៖ គណនា​ផលបូក 1/u1+1/u2+…

តាង u_0,u_1,u_2,\dotsi,u_n  ជាស្វ៊ីត​នៃ​ចំនួន​ពិត ដែល​បំពេញ​ទំនាក់​ទំនង​ខាង​ក្រោម​៖

(3-u_{n+1})(6+u_n)=18  និង  u_0=3

គណនា

\displaystyle \sum_{k=0}^{2011}{\frac{1}{u_k}}

ចម្លើយ

តាង \displaystyle t_k=\frac{1}{u_k} ។ ដូច្នេះ \displaystyle t_0=\frac{1}{3} និង

\displaystyle \left(3-\frac{1}{t_{n+1}}\right)\left(6+\frac{1}{t_{n}}\right)=18
\displaystyle \Leftrightarrow \frac{3}{t_{n}}-\frac{6}{t_{n+1}}-\frac{1}{t_nt_{n+1}}=0
\displaystyle \Leftrightarrow 3t_{n+1}-6t_n-1=0 (*)
\displaystyle \Leftrightarrow t_{n+1}=2t_n+\frac{1}{3}

ដូច្នេះ

\displaystyle t_1=2t_0+\frac{1}{3} …….×2^{n-1}
\displaystyle t_2=2t_1+\frac{1}{3} …….×2^{n-2}

\displaystyle t_{n-2}=2t_{n-3}+\frac{1}{3} …….×2^2
\displaystyle t_{n-1}=2t_{n-2}+\frac{1}{3} …….×2^1
\displaystyle t_n=2t_{n-1}+\frac{1}{3} …….×2^0

គុណសមភាពទី១ ទី២ …. ទី (n-2) ទី (n-1) និង ទី n នឹង 2^{n-1},2^{n-2},\dotsi,2^2,2^1,2^0 រួចបូកសមភាពទទួលបានបញ្ចូលគ្នា យើងទទួលបាន

\displaystyle t_n=2^n t_0+\frac{1}{3}\left(1+2+2^2+\dotsi +2^{n-1}\right)
\displaystyle =\frac{1}{3}.2^n+\frac{1}{3}\left(1+2+2^2+\dotsi +2^{n-1}\right)
\displaystyle =\frac{1}{3}\left(1+2+2^2+\dotsi +2^{n-1}+2^n\right)
\displaystyle =\frac{1}{3}\frac{2^{n+1}-1}{2-1}=\frac{1}{3}\left(2^{n+1}-1\right)

ម្យ៉ាងវិញទៀត តាមសមភាព (*) ដោយ យក n=0,1,2,\dotsi យើងទទួលបាន

3\left(t_1-t_0\right)=3t_0+1
3\left(t_2-t_1\right)=3t_1+1
3\left(t_3-t_2\right)=3t_2+1

3\left(t_{n+1}-t_n\right)=3t_n+1

បូកសមភាពទាំងនេះបញ្ចូលគ្នា យើងទទួលបាន

\displaystyle 3S_n+n+1=3\left(t_{n+1}-t_0\right)
\displaystyle \implies S_n=\frac{2^{n+2}-n-3}{3}

ដែល

\displaystyle S_n=t_0+t_1+t_2+\dotsi+t_n=\sum_{k=0}^{2011}{\frac{1}{u_k}}

យក n=2011  យើងទទួលបាន

\displaystyle \sum_{k=0}^{2011}{\frac{1}{u_k}}=\frac{2^{2013}-2014}{3}

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