Let $$\{a_n\}_{n=1}^\infty$$ 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 ___.

Let

$$G(x)=\sum_{n=1}^{\infty}a_nx^n$$

Then,

$$G(x)=x+x^2+\sum_{n=1}^{\infty}a_{n+2}x^{n+2}$$

Using

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

we get

$$G(x)=x+x^2+\sum_{n=1}^{\infty}(2a_{n+1}+a_n)x^{n+2}$$ $$=x+x^2+2x\sum_{n=1}^{\infty}a_{n+1}x^{n+1}+x^2\sum_{n=1}^{\infty}a_nx^n$$

Now,

$$\sum_{n=1}^{\infty}a_nx^n=G(x)$$ and $$\sum_{n=1}^{\infty}a_{n+1}x^{n+1}=G(x)-x$$

Substituting,

$$G(x)=x+x^2+2x(G(x)-x)+x^2G(x)$$

$$G(x)=x+x^2+2xG(x)-2x^2+x^2G(x)$$

$$G(x)(1-2x-x^2)=x-x^2$$

Therefore,

$$G(x)=\frac{x-x^2}{1-2x-x^2}$$

Now,

$$\sum_{n=1}^{\infty}\frac{a_n}{2^{3n}}=G\left(\frac18\right)$$

So,

$$G\left(\frac18\right)=\frac{\frac18-\frac1{64}}{1-\frac14-\frac1{64}}$$

$$=\frac{\frac7{64}}{\frac{47}{64}}$$

$$=\frac7{47}$$

Hence,

$$47\sum_{n=1}^{\infty}\frac{a_n}{2^{3n}}=47\cdot\frac7{47}=7$$