One more sum (edited)

$\begin{aligned} S & = \frac 12 + \frac 54 + \frac {14}8 + \frac {30}{16} + \cdots \\ & = \frac {1^2}2 + \frac {1^2+2^2}{2^2} + \frac {1^2+2^2+3^2}{2^3} + \frac {1^2+2^2+3^2+4^2}{2^4} + \cdots \\ & = \small \frac {1^2}2 \left(1+\frac 12 + \frac 1{2^2} + \cdots\right) + \frac {2^2}{2^2} \left(1+\frac 12 + \frac 1{2^2} + \cdots\right) + \frac {3^2}{2^3} \left(1+\frac 12 + \frac 1{2^2} + \cdots\right) + \cdots \\ & = \left(\frac {1^2}2 + \frac {2^2}{2^2} + \frac {3^2}{2^3} + \cdots\right)\blue{\left(\frac 1{1-\frac 12} \right)} \\ & = \blue 2 \sum_{n=1}^\infty \frac {n^2}{2^n} = 2 \sum_{\red{n=0}}^\infty \frac {n^2}{2^n} = 2 \sum_{\red{n=0}}^\infty \frac {(n+1)^2}{2^{n+1}} \end{aligned}$

$\begin{aligned} \ \ & = \sum_{\red{n=0}}^\infty \frac {n^2+2n+1}{2^n} \\ & = \blue{\sum_{n=0}^\infty \frac {n^2}{2^n}} + 2 \red{\sum_{n=0}^\infty \frac n{2^n}} + \sum_{n=0}^\infty \frac 1{2^n} & \small \blue{\text{Note that }S = 2\sum_{n=0}^\infty \frac {n^2}{2^n}} \\ & = \blue{\frac 12 S} + 2 \red{\sum_{n=0}^\infty \frac {n+1}{2^{n+1}}} + \frac 1{1-\frac 12} & \small \red{\text{Using the same method}} \\ & = \frac 12 S + \sum_{n=0}^\infty \frac n{2^n} + \sum_{n=0}^\infty \frac 1{2^n} + 2 \\ & = \frac 12 S + 2(2) + 2 \\ & = 12 \end{aligned}$

Therefore, $p+q = 12+1 = \boxed{13}$ .

The infinite sum can be rewritten as follows:

S = $\displaystyle \sum_{n=1}^\infty$ $\frac {\sum_{k=1}^n k²}{2^n}$

= $\displaystyle \sum_{n=1}^\infty$ $\displaystyle\frac {\frac{ n(n+1)(2n+1)}{6}}{2^n}$

= $\displaystyle \sum_{n=1}^\infty$ $\displaystyle\frac {\frac{n³}{3} + \frac{n²}{2} + \frac{n}{6}}{2^n}$

= $\frac {\displaystyle \sum_{n=1}^\infty \frac {n³}{2^n}}{\displaystyle 3}$ + $\frac {\displaystyle \sum_{n=1}^\infty \frac {n²}{2^n}}{\displaystyle 2}$ + $\frac {\displaystyle \sum_{n=1}^\infty \frac {n}{2^n}}{\displaystyle 6}$

= $\frac{26}{3} + \frac{6}{2} + \frac{2}{6}$

= $\displaystyle\frac{12}{1}$

Therefore $p = 12 , q = 1$

$p + q = 13$

Additional information:

Value of $\displaystyle \sum_{n=1}^\infty \frac {n³}{2^n}$

Let K = $\displaystyle \sum_{n=1}^\infty \frac {n³}{2^n}$

= $\displaystyle \sum_{n=0}^\infty \frac {(n+1)³}{2^{n+1}}$

= $\displaystyle \sum_{n=0}^\infty \frac {(n³ + 3n² + 3n + 1)}{2^{n+1}}$

= $\displaystyle \frac {\sum_{n=0}^\infty \frac {n³}{2^n}}{\displaystyle 2}$ + 3 $\displaystyle \frac {\sum_{n=0}^\infty \frac {n²}{2^n}}{\displaystyle 2}$ + 3 $\displaystyle \frac {\sum_{n=0}^\infty \frac {n}{2^n}}{\displaystyle 2}$ + $\displaystyle \frac {\sum_{n=0}^\infty \frac {1}{2^n}}{\displaystyle 2}$

= $\frac{K}{2}$ + 3 $\displaystyle \frac {\sum_{n=1}^\infty \displaystyle\frac {n²}{2^n}}{\displaystyle 2}$ + 3 $\displaystyle \frac {\sum_{n=1}^\infty \displaystyle\frac {n}{2^n}}{\displaystyle 2}$ + $\displaystyle \frac {\sum_{n=0}^\infty \displaystyle\frac {1}{2^n}}{\displaystyle 2}$

On subtracting $\frac{K}{2}$ on both sides and multiplying the resultants by 2 we get

K = 3 $\displaystyle \sum_{n=1}^\infty \frac {n²}{2^n}$ + 3 $\displaystyle \sum_{n=1}^\infty \frac {n}{2^n}$ + $\displaystyle \sum_{n=0}^\infty \frac {1}{2^n}$

= $\ 3×6 + 3×2 + 2$

$= 26$

Sorry for the errors in the previous version