Summation this Summer! 2

$\text{Let's start with the following series : }$

$\large\displaystyle \mathfrak{S} = \displaystyle \sum_{k=1}^{\infty} \frac{k^2}{2^k}$

$\large\displaystyle \implies \mathfrak{S} = \frac{1^2}{2} + \frac{2^2}{2^2} + \frac{3^2}{2^3} + \cdots \infty \space \text{(1)}$

$\large \displaystyle\implies \frac{\mathfrak{S}}{2} = \frac{1^2}{2^2} + \frac{2^2}{2^3} + \cdots \infty \space \text{(2)}$

$\text{Subtracting (2) from (1)}$

$\large \displaystyle\implies \frac{\mathfrak{S}}{2} = \frac{1}{2} + \frac{2^2-1^2}{2^2} + \frac{3^2-2^2}{2^3} + \cdots \infty \space \text{(3)}$

$\large\displaystyle \implies \frac{\mathfrak{S}}{4} =\frac{1}{2^2}+ \frac{3}{2^3} + \frac{5}{2^4} + \cdots \infty \space \text{(4)}$

$\text{Subtracting (4) from (3)}$

$\large\displaystyle \implies \frac{\mathfrak{S}}{4} = \frac{1}{2} + \frac{3-1}{2^2} + \frac{5-3}{2^3} + \cdots \infty$

$\large \displaystyle\implies \frac{\mathfrak{S}}{4} = \frac{1}{2} + \frac{2}{2^2}\color{#D61F06}{\underbrace{(1 + \frac{1}{2} + \frac{1}{2^2} + \cdots \infty)}}$

$\text{The above braced series is an }$ Infinite Geometric Progression $\text{ with common ratio }\frac{1}{2}$

$\large\displaystyle \implies \frac{\mathfrak{S}}{4} = \frac{1}{2} + \frac{2}{2^2}\frac{1}{1-\frac{1}{2}}$

$\large\displaystyle \implies \frac{\mathfrak{S}}{4} = \frac{3}{2}\implies \mathfrak{S} = 6$

$\text{The series in the problem can be written as :}$

$\large\displaystyle \mathfrak{A} = \sum_{k=1}^{\infty} \frac{(2k-1)(2k+1)}{2^k}$

$\large\displaystyle \mathfrak{A} = \sum_{k=1}^{\infty} \frac{4k^2-1}{2^k}$

$\large \displaystyle\implies \mathfrak{A} = 4\underbrace{\sum_{k=1}^{\infty}\frac{k^2}{2^k}} - \underbrace{\sum_{k=1}^{\infty} \frac{1}{2^k}}$

$\large \displaystyle\implies \mathfrak{A} = 4 .\underbrace{\mathfrak{S}} - \underbrace{\frac{\frac{1}{2}} {1-\frac{1}{2}}}$

$\large \displaystyle\implies \mathfrak{A} = 24 - 1\implies \boxed{\mathfrak{A} = 23}$

Here I will use generating function:
Let $f(x)=1\times 3x+3\times 5x^2+5\times 7x^3+7\times 9x^4+\cdots$
So we have to find $f(\dfrac 12)$
$\begin{array}{c}\\f(x)(1-x)&=1\times 3x&+4\times 3x^2&+4\times 5x^3&+4\times 7x^4&+\cdots\\f(x)(1-x)^2&=1\times 3x&+3\times 3x^2&+4\times 2x^3&+4\times 2x^4&+\cdots\\&=3x&+9x^2&+8x^3&+8x^4&+\cdots\\f(x)(1-x)^3&=3x&+6x^2&-x^3\end{array}$
$\therefore f(x)=\dfrac{3x+6x^2-x^3}{(1-x)^3}$
Substitute $x=\dfrac 12$ yields $f(\dfrac 12)=23$

I have a slightly different solution. The sequence we sum is $\frac{(2n-1)(2n+1)}{2^n}$ . The denominator is growing as $x^n$ when $x=1/2$ , which calls for the use of generating functions. The sequence $a_n=(2n-1)(2n+1)=4n^2-1$ is generated by $F(x)=4(xD)^2(\frac{x}{1-x})-\frac{x}{1-x}=\frac{4(x^2+x)}{(1-x)^3}-\frac{x}{1-x}$ . Substituting $x=1/2$ gives the result.