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

Let, $S={\sum }_{n=1}^{\infty }\frac{a_n}{2^{3n}}$

$64\frac{a_{n+2}}{2^{3(n+2)}}=16\frac{a_{n+1}}{2^{3(n+1)}}+\frac{a_n}{2^{3n}}$

Putting $n=1,2,3,\mathrm{...}$ and adding,

$\begin{array}{l}64(S-\frac{a_1}{2^3}-\frac{a_2}{2^6})=16(S-\frac{a_1}{2^3})+S\\ \Rightarrow 47S=6a_1+a_2=8.\end{array}$