Do you know it?

We know for $|x|<1$

$\large{\displaystyle \sum_{n=1}^{\infty} x^n=\frac{1}{1-x}}$

on differentiating with respect to x

$\large{\displaystyle \sum_{n=1}^{\infty} nx^{n-1}=\frac{1}{(1-x)^2}}$

$\large{\displaystyle \sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}}$

on differentiating again and then multiplying with $x$ we get

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

differentiating again

$\large{\displaystyle \sum_{n=1}^{\infty} n^3 x^{n-1}=\frac{(1-x)^3(1+2x)+3(x+x^2)(1-x)^2}{(1-x)^6}}$

substitute $x=\frac{1}{2}$

we get $\large{\boxed{52}}$

$\displaystyle\sum_{n=1}^{\infty}\frac{n^3}{2^{n-1}}=\frac{d^3}{dx^3}\sum_{n=1}^{\infty}\frac{e^{nx}}{2^{n-1}}\bigg|_{x=0}$

$\displaystyle=2\frac{d^3}{dx^3}\sum_{n=1}^{\infty}\left(\frac{e^x}{2}\right)^n\bigg|_{x=0}=2\frac{d^3}{dx^3}\frac{e^x}{2-e^x}\bigg|_{x=0}$

$\displaystyle=\frac{4e^{3x}+32e^{2x}+16e^x}{(2-e^x)^4}\bigg|_{x=0}$

$\displaystyle=\frac{4+32+16}{(2-1)^4}=\boxed{52}$ .

$\displaystyle \begin{aligned} (1-x)^{-1} &= 1 + x + x^2 + x^3 + \cdots\\ \text{Differentiating }\Rightarrow x(1-x)^{-2} &= x + 2x^2 + 3x^3 + 4x^4 + \cdots\\ \small\text{Differentiating again}\\ \normalsize\Rightarrow x \left(2x(1-x)^{-3} + (1-x)^{-2} \right) &= x + 2^2 x^2 + 3^2 x^3 + 4^2 x^4 + \cdots\\ &=2 x^2(1-x)^{-3} + x(1-x)^{-2} \\ \text{Differentiating } \\ \Rightarrow 4x(1-x)^{-3} + 6x^2(1-x)^{-4} \nonumber \\ + (1-x)^{-2} + 2x(1-x)^{-3} &= 1 + 2^3x + 3^3x^2 + 4^3x^3 + \cdots\\\\ 4x(1-x)^{-3} + 6x^2(1-x)^{-4} \nonumber \\ + (1-x)^{-2} + 2x(1-x)^{-3} &=\frac{4x(1-x) + 6x^2 + (1-x)^2 + 2x(1-x)}{(1-x)^4}\\ &=\frac{x^2+4x+1}{(1-x)^4}\end{aligned}$

Substituting $x = 0.5$

$\frac{x^2+4x+1}{(1-x)^4} = \left(\frac{1}{4} + 3\right) \times 16$

$\boxed{\therefore \text{The sum } = 52}$

Poly logarithm makes it easy!!!