Question

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

Answer 1

$a_{n+2}=2 a_{n+1}+a_{n^{3}}$
let $\displaystyle\sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}}=P$
Divide by $8^{n}$ we get
$\frac{a_{n+2}}{8^{n}}=\frac{2 a_{n+1}}{8^{n}}+\frac{a_{n}}{8^{n}}$
$\Rightarrow 64 \frac{a_{n+2}}{8^{n+2}}=\frac{16 a_{n+1}}{8^{n+1}}+\frac{a_{n}}{8^{n}} $
$64 \displaystyle\sum_{n=1}^{\infty} \frac{a_{n+2}}{8^{n+2}}=16 \displaystyle\sum_{n=1}^{\infty} \frac{a_{n+1}}{8^{n+1}}+\displaystyle\sum_{n=1}^{\infty} \frac{a_{n}}{8^{n}} $
$64\left(P-\frac{a_{1}}{8}-\frac{a_{2}}{8^{2}}\right)=16\left(P-\frac{a_{1}}{8}\right)+P $
$\Rightarrow 64\left(P-\frac{1}{8}-\frac{1}{64}\right)=16\left(P-\frac{1}{8}\right)+P$
$64 P-8-1=16 P-2+P $
$47 P=7$