Question

Let ${\left\{a_n\right\}}_{n=1}^{\infty }$ be a sequence such that $a_1=1,a_2=1$ and $a_{n+2}=2a_{n+1}+a_n$ for all $n\ge 1$. Then the value of $47\sum _{n=1}^{\infty }\frac{a_n}{2^{3n}}$ is equal to_______.

Answer 1

Answer: 7

$a_{n+2}=2a_{n+1}+a_{n^3}$
let $\sum _{n=1}^{\infty }\frac{a_n}{8^n}=P$
Divide by $8^n$ we get
$\frac{a_{n+2}}{8^n}=\frac{2a_{n+1}}{8^n}+\frac{a_n}{8^n}$
$\Rightarrow 64\frac{a_{n+2}}{8^{n+2}}=\frac{16a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}$
$64\sum _{n=1}^{\infty }\frac{a_{n+2}}{8^{n+2}}=16\sum _{n=1}^{\infty }\frac{a_{n+1}}{8^{n+1}}+\sum _{n=1}^{\infty }\frac{a_n}{8^n}$
$64\left(P-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(P-\frac{a_1}{8}\right)+P$
$\Rightarrow 64\left(P-\frac{1}{8}-\frac{1}{64}\right)=16\left(P-\frac{1}{8}\right)+P$
$64P-8-1=16P-2+P$
$47P=7$