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 $47\sum _{n=1}^{\infty }\left(\frac{a_n}{2^{3n}}\right)$ is equal to ________.

We have,

$47\sum _{n=1}^{\infty }\left(\frac{a_n}{2^{3n}}\right)$

$\Rightarrow \sum _{n=1}^{\infty }\left(\frac{a_n}{8^n}\right)=P$(let)

And,

$a_{n+2}=2a_{n+1}+a_n$

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

$\Rightarrow 64\left(\frac{a_{n+2}}{8^{n+2}}\right)=\frac{16a_{n+1}}{8^{n+1}}+\frac{a_n}{8^n}$

$\Rightarrow 64\sum _{n=1}^{\infty }\left(\frac{a_{n+2}}{8^{n+2}}\right)=16\sum _{n=1}^{\infty }\left(\frac{a_{n+1}}{8^{n+1}}\right)+\sum _{n=1}^{\infty }\left(\frac{a_n}{8^n}\right)$

$\Rightarrow 64\left(P-\frac{a_1}{8}-\frac{a_2}{8^2}\right)=16\left(P-\frac{a_1}{8}\right)+P$

$\Rightarrow 64\left(P-\frac{1}{8}-\frac{1}{64}\right)=16\left(P-\frac{1}{8}\right)+P$

$\Rightarrow 64P-8-1=16P-2+P$

$\Rightarrow 47P=7$

$\Rightarrow 47\sum _{n=1}^{\infty }\left(\frac{a_n}{8^n}\right)=7$