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माना कि $x ^2- x -1=0$ के मूल (roots) $\alpha$ और $\beta$ हैं, जहाँ $\alpha>\beta$ है। सभी धनात्मक पूर्णांकों $n$ के लिए निम्न को परिभाषित किया गया है
$a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}, n \geq 1$
$b_1=1 \text { and } b_n=a_{n-1}+a_{n+1}, n \geq 2.$
तब निम्न में से कौनसा (से) विकल्प सही है (हैं) ?
$(1)$ प्रत्येक $n \geq 1$ के लिए, $a _1+ a _2+ a _3+\ldots . .+ a _{ n }= a _{ n +2}-1$
$(2)$ $\sum_{ n =1}^{\infty} \frac{ a _{ n }}{10^{ n }}=\frac{10}{89}$
$(3)$ $\sum_{ n =1}^{\infty} \frac{ b _{ n }}{10^{ n }}=\frac{8}{89}$
$(4)$ प्रत्येक $n \geq 1$ के लिए, $b _{ n }=\alpha^{ n }+\beta^{ n }$
$1,2,3$
$1,2$
$1,2,4$
$2,3$
Solution
$\alpha, \beta$ are roots of $x^2-x-1$
$a_{r+2}-a_t=\frac{\left(\alpha^{x+2}-\beta^{r+2}\right)-\left(\alpha^x-\beta^r\right)}{\alpha-\beta}=\frac{\left(\alpha^{x+2}-\alpha^x\right)-\left(\beta^{x+2}-\beta^x\right)}{\alpha-\beta}$
$=\frac{\alpha^x\left(\alpha^2-1\right)-\beta^x\left(\beta^2-1\right)}{\alpha-\beta}=\frac{\alpha^T \alpha-\beta^T \beta}{\alpha-\beta}=\frac{\alpha^{r+1}-\beta^{T+1}}{\alpha-\beta}=a_{r+1}$
$\Rightarrow a_{x+2}-a_{r+1}=a_t$
$\Rightarrow \sum_{x=1}^n a_x=a_{n+2}-a_2=a_{n+2}-\frac{\alpha^2-\beta^2}{\alpha-\beta}=a_{n+2}-(\alpha+\beta)=a_{n+2}-1$
$\text { Now } \sum_{n=1}^{\infty} \frac{a_n}{10^n}=\frac{\sum_{n=1}^x\left(\frac{\alpha}{10}\right)^n-\sum_{n=1}^x\left(\frac{\beta}{10}\right)^n}{\alpha-\beta}$
$=\frac{\frac{\alpha}{10}}{1-\frac{\alpha}{10}}-\frac{\frac{\beta}{10}}{\alpha-\frac{\beta}{10}}=\frac{\frac{\alpha}{10-\alpha}-\frac{\beta}{10-\beta}}{(\alpha-\beta)}=\frac{10}{(10-\alpha)(10-\beta)}=\frac{10}{89}$
$\sum_{n=1}^{\infty} \frac{b_n}{10^n}=\sum_{n=1}^{\infty} \frac{a_{n-1}+a_{n+1}}{10^n}=\frac{\frac{\alpha}{10}}{1-\frac{\alpha}{10}}+\frac{\frac{\beta}{10}}{1-\frac{\beta}{10}}=\frac{12}{89}$
Further, $b_n=a_{n-1}+a_{n+1}$
$=\frac{\left(\alpha^{2=1}-\beta^{2-1}\right)+\left(\alpha^{2+1}-\beta^{2+1}\right)}{\alpha-\beta}$
$\text { (as } \left.\alpha \beta=-1 \Rightarrow \alpha^{n-1}=-\alpha^n \beta \& \beta^{n-1}=-\alpha \beta^n\right)$
$=\frac{\alpha^2(\alpha-\beta)+(\alpha-\beta) \beta^n}{\alpha-\beta}=\alpha^n+\beta^n$
