Infinite Sum

We can also use AGP.

$\Rightarrow 1\left(\dfrac{1}{2}\right)^0+2\left(\dfrac{1}{2}\right)^1+3\left(\dfrac{1}{2}\right)^2+4\left(\dfrac{1}{2}\right)^3+5\left(\dfrac{1}{2}\right)^4+...$

$\implies 1+\dfrac{2}{2}+\dfrac{3}{4}+\dfrac{4}{8}+\dfrac{5}{16}+...$

Here: $a=1$ , $r=\dfrac{1}{2}$ and $d=1$ .

We know that: $S_{\infty}=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}$ .

$\implies \dfrac{1}{1-\frac{1}{2}}+\dfrac{1×\frac{1}{2}}{\left(1-\frac{1}{2}\right)^2}$

$\implies 2+4=\boxed{6}$

$\dfrac{1}{1-x} = \displaystyle \sum_{n=0}^{\infty} x^{n} , |x| < 1$
Differentiating w.r.t x,
$\dfrac{1}{(1-x)^2} = \displaystyle \sum_{n=0}^{\infty} nx^{n-1}$
Put $x = \dfrac{1}{2}$
$\displaystyle \sum_{n=0}^{\infty} \dfrac{n}{2^{n-1}} = 4$