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Consider for $|x|<1$ :

$\sum_{n=0}^{\infty} x^n= \dfrac{1}{1-x}$ Differentiate w.r.t $x$ and multiply by $x$ on both sides:

$\sum_{n=0}^{\infty} nx^n=\dfrac{x}{(1-x)^2}=\dfrac{1}{(1-x)^2}-\dfrac{1}{1-x}$

Differentiate w.r.t $x$ and multiply by $x$ on both sides:

$\begin{aligned}\sum_{n=0}^{\infty} n^2 x^n=&\dfrac{2x}{(1-x)^3}-\dfrac{x}{(1-x)^2}\\=&\dfrac{2}{(1-x)^3}-\dfrac{3}{(1-x)^2}+\dfrac{1}{1-x}\end{aligned}$

Differentiate w.r.t $x$ and multiply by $x$ on both sides:

$\sum_{n=0}^{\infty}n^3 x^n=\dfrac{6x}{(1-x)^4}-\dfrac{6x}{(1-x)^3}+\dfrac{x}{(1-x)^2}$

Put $x=\dfrac 12$ to get:

$\displaystyle \sum_{n=0}^{\infty} \dfrac {n^3}{2^n}=\boxed{\color{#D61F06}26}$

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$\Large \color{cyan}{\mathcal{Alternate~Method}}$

$\begin{aligned}\mathfrak Z= \displaystyle \sum_{n=0}^{\infty} \dfrac {n^3}{2^n}=\dfrac{1}{2}+&\dfrac{8}{2^2}+\dfrac{27}{2^3}+\dfrac{64}{2^4}+\cdots\\\dfrac{\mathfrak Z}{2}=~~~~~~~~~~~~~~~~~~~~~~~~&\dfrac{1}{2^2}+\dfrac{8}{2^3}+\dfrac{27}{2^4}+\cdots\end{aligned}$

Subtracting we get:

$\begin{aligned}\dfrac{\mathfrak Z}{2}=\dfrac{1}{2}+&\dfrac{7}{2^2}+\dfrac{19}{2^3}+\dfrac{37}{2^4}+\cdots\\\dfrac{\mathfrak Z}{4}=~~~~~~~~&\dfrac{1}{2^2}+\dfrac{7}{2^3}+\dfrac{19}{2^4}+\cdots\end{aligned}$

Subtracting we get:

$\begin{aligned}\dfrac{\mathfrak Z}{4}=\dfrac{1}{2}+&\dfrac{6}{2^2}+\dfrac{12}{2^3}+\dfrac{18}{2^4}+\cdots\\\dfrac{\mathfrak Z}{8}=~~~~~~~~~~~&\dfrac{1}{2^2}+\dfrac{6}{2^3}+\dfrac{12}{2^4}+\cdots \end{aligned}$

Subtracting we get:

$\dfrac{\mathfrak Z}{8}=\dfrac{1}{2}+\dfrac{5}{2^2}+\dfrac{6}{2^3}+\dfrac{6}{2^4}\cdots$ $\begin{aligned}\implies \mathfrak Z=&4+10+6\overbrace{\left[1+\dfrac{1}{2}+\dfrac{1}{4}+\cdots\right]}^{\color{#3D99F6}{\text{Infinite GP}}}\\=&4+10+6(2)=\boxed{\color{#D61F06}26}\end{aligned}$

Consider $\sum_{n=0}^\infty\frac{n^{3}}{2^{n}}$ .

Since the first term is 0, it can be written as $\sum_{n=1}^\infty\frac{n^{3}}{2^{n}}$ , or $\sum_{n=0}^\infty\frac{(n+1)^{3}}{2^{n+1}}$ .

Expanding $(n+1)^{3}$ , it can be written as $\frac{\sum_{n=0}^\infty\frac{n^{3}}{2^{n}} + 3\sum_{n=0}^\infty\frac{n^{2}}{2^{n}} + 3\sum_{n=0}^\infty\frac{n}{2^{n}} + \sum_{n=0}^\infty\frac{1}{2^{n}}}{2}$ .

It simplifies to $2 + 3\sum_{n=0}^\infty\frac{n^{2}}{2^{n}} + 3\sum_{n=0}^\infty\frac{n}{2^{n}}$ .

Similarly, $\sum_{n=0}^\infty\frac{n^{2}}{2^{n}} = 2 + 2\sum_{n=0}^\infty\frac{n}{2^{n}}$ .

Further, taking $\sum_{n=0}^\infty\frac{n}{2^{n}}$ as $\sum_{n=1}^\infty\frac{n}{2^{n}}$ and so on, we get $\sum_{n=0}^\infty\frac{n}{2^{n}} = 2$ .

Thus by substituting, we get $\sum_{n=0}^\infty\frac{n^{3}}{2^{n}} = 26$ .