Sum of Infinite series $frac{1.3}{2}+frac{3.5}{2^2}+frac{5.7}{2^3}+frac{7.9}{2^4}+......$

Prove that the sum of the infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ is 23.

My approach
I got the following term

$S_n=\sum_1^\infty\frac{4n^2}{2^n}-\sum_1^\infty\frac{1}{2^n}$.

For $\sum_1^\infty\frac{1}{2^n}$ the answer is 1 as it forms a geometric series but I am bot able to find the solution to $\sum_1^\infty\frac{4n^2}{2^n}$.

Answer

$$\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+…=\sum_{n=1}^{\infty }\frac{(2n-1)(2n+1)}{2^n}=\sum_{n=1}^{\infty }\frac{(4n^2-1)}{2^n}$$
depending on the geometric series
$$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
$$(\frac{1}{1-x})'=\sum_{n=1}^{\infty }nx^{n-1}$$
$$x(\frac{1}{1-x})'=\sum_{n=1}^{\infty }nx^{n}$$
$$(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }n^2x^{n-1}$$

$$x(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }n^2x^{n}$$
$$4x(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }4n^2x^{n}$$
$$4x(x(\frac{1}{1-x})')'-\frac{1}{1-x}=\sum_{n=1}^{\infty }4n^2x^{n}-\sum_{n=0}^{\infty }x^n$$
$$4x(x(\frac{1}{1-x})')'-\frac{1}{1-x}+1=\sum_{n=1}^{\infty }4n^2x^{n}-\sum_{n=1}^{\infty }x^n$$
so
$$\sum_{n=1}^{\infty }x^n(4n^2-1)=\frac{4x^2+4x}{(1-x)^3}-\frac{x}{1-x}$$
now let $x=1/2$ to get $23$