Infinity and Beyond

Let the given sum be S.

$S = \displaystyle \sum_{n=1}^{\infty} \dfrac{(n)(n+1)}{2^{n-1}} = \displaystyle \sum_{1}^{\infty} \dfrac{n^{2} + n}{2^{n-1}}$

Absolute convergence can be proved using the ratio test, and hence we are free to rearrange the terms.

Let, $S_{1} = \displaystyle \sum_{n=1}^{\infty} \dfrac{n}{2^{n-1}}$

$S_{1} = 1 + \dfrac{2}{2} + \dfrac{3}{4} + \ldots$ ...(1)

$\dfrac{S_{1}}{2} = \dfrac{1}{2} + \dfrac{2}{4} + \dfrac{3}{8} + \ldots$ ...(2)
Subtracting (2) from (1),
$\dfrac{S_{1}}{2} = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} \ldots$
$S_{1} = 2\cdot \dfrac{1}{1-\dfrac{1}{2}} = 4$

Let $S_{2} = \displaystyle \sum_{n=1}^{\infty} \dfrac{n^{2}}{2^{n-1}}$

$S_{2} = 1 + \dfrac{4}{2} + \dfrac{9}{4} + \ldots$ ..(3)
$\dfrac{S_{2}}{2} = \dfrac{1}{2} + \dfrac{4}{4} + \dfrac{9}{8} \ldots$ ...(4)
Subtracting (4) from (3),
$\dfrac{S_{2}}{2} = 1 + \dfrac{3}{2} + \dfrac{5}{4} + \ldots$ ...(5)

$\dfrac{S_{2}}{4} = \dfrac{1}{2} + \dfrac{3}{4} + \dfrac{5}{8} + \ldots$ ...(6)
Subtracting (6) from (5),
$\dfrac{S_{4}}{4} = 1 + \dfrac{2}{2} + \dfrac{2}{4} + \dfrac{2}{8} + \ldots$
$\dfrac{S_{4}}{4} = 1 + \dfrac{1}{1-\dfrac{1}{2}} = 3$
$\therefore S_{2} = 4\cdot 3 = 12$
$S = S_{1} + S_{2} = 4 + 12 = 16$

This is calculus approach

This sum can be rewritten as :

$\frac{2.1}{2^{0}}+\frac{3.2}{2}+\frac{4.3}{2^{2}}+\frac{5.4}{2^{3}}+... = \sum_{n=0}^{\infty}\frac{(n+2)(n+1)}{2^{n}} = \sum_{n=2}^{\infty}\frac{n(n-1)}{2^{n-2}}=\sum_{n=2}^{\infty}n(n-1)x^{n-2}$ where $x=\frac{1}{2}$

This last series can be summed by differentiating the geometric series twice.

So , for any $\left | x \right |<1$ we have :

$f(x)=\sum_{n=0}^{\infty }x^{n}=\frac{1}{1-x}$

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

$f''(x)=\sum_{n=0}^{\infty }n(n-1)x^{n-2}=\frac{2}{(1-x)^{3}}$

Evaluating this last expression at $x = 1/2$ , and seeing that this is the same as the series we need to sum, we have the sum which is 16.

I have solved this by using arithmetic optometric sequence twice