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  • Let   be a sequence such that <img alt="a_{1}=1, a_{2}=1" src="/latex-image/?a_%7B1%7D%3D1%2C%20a_%7B2%7D%3D1"

Let \left\{a_{n}\right\}_{n=1}^{\infty}  be a sequence such that a_{1}=1, a_{2}=1 and \mathrm{a}_{\mathrm{n}+2}=2 \mathrm{a}_{\mathrm{n}+1}+\mathrm{a}_{\mathrm{a}}$ for all $\mathrm{n} \geq 1. Then the value of 47 \sum_{n=1}^{\infty} \frac{a_{n}}{2^{3 n}}is equal to_________.
 

Answers (1)

best_answer

\\a_{1}=1, a_{2}=1\; and\\ a_{n+2}=2 a_{n+1}+a_{n}       ........(i)

We need to find \sum_{n=1}^{\infty} \frac{a_{n}}{2^{3 n}}=\sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}

\text { Let } \sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}=S

Now (i) :

\begin{aligned} &a_{n+2}=2 \cdot a_{n+1}+a_{n} \\ &\Rightarrow \frac{a_{n+2}}{8^{n}}=2 \cdot \frac{a_{n+1}}{8^{n}}+\frac{a_{n}}{8^{n}} \end{aligned}

\Rightarrow 8^{2} \frac{a_{n+2}}{8^{n+2}}=2.8 \frac{a_{n+1}}{8^{n+1}}+\frac{a_{n}}{8^{n}}

Apply sunmation from n=1 \text { till } \infty \text {. }

\begin{aligned} &\Rightarrow 64 \sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}}=16 \sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}}+\sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}} \\ &\Rightarrow 64\left[S-\frac{a_{1}}{8}-\frac{a_{2}}{8^{2}}\right]=16\left(s-\frac{a_{1}}{8}\right)+S \end{aligned}

\\\Rightarrow \quad Put \; a_{1}=a_{2}=1\\ \\\Rightarrow S=\frac{7}{47}

\Rightarrow 47\; S=7

Hence, the correct answer is (7).

Posted by

Kuldeep Maurya

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