Triangulate the answer

$S=\frac{1}{2}+\frac{3}{4}+\frac{6}{8}+\frac{10}{16}+\frac{15}{32}+...$ $\frac{S}{2}=\hspace{9 mm}\frac{1}{4}+\frac{3}{8}+\frac{6}{16}+\frac{10}{32}+...$

Subtracting,

$\frac{S}{2}=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+...$ $\frac{S}{4}=\hspace{9 mm}\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+...$

Again subtracting,

$\frac{S}{4}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$

Its an infinite GP with sum=1 $\Big (\dfrac{\frac{1}{2}}{1-\frac{1}{2}}\Big )$

So $\boxed{S=4}$

It is known that $\displaystyle \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ for $|x| < 1$ .

Taking the second derivative on both sides, we obtain

$\frac{d^{2}}{dx^{2}} \sum_{n=0}^{\infty} x^{n} = \sum_{n=0}^{\infty} n(n-1)x^{n-2} = \sum_{n=1}^{\infty} n(n+1)x^{n-1}$

for the LHS, and

$\frac{d^{2}}{dx^{2}} \frac{1}{1-x} = \frac{2}{(1-x)^{3}}$

for the RHS after repeated chain rule.

Combining the LHS and RHS and dividing both sides by 4 gives us

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

It is also well known that the nth triangular number $T_{n} = 1+2+\ldots+n = \frac{n(n+1)}{2}$ . Using this fact and substituting in $x = \frac{1}{2}$ , we get

$\sum_{n=1}^{\infty} \frac{T_{n}}{2^{n}} = \frac{1}{2(1-\frac{1}{2})^{3}} = \boxed{4}$

let's first find this sum $\Sigma_{n=1}^\infty 2^{-an} = \frac{1}{2^{a}-1}$

then we differentiate the equation $\frac{\partial}{\partial a} (\Sigma_{n=1}^\infty 2^{-an})= \frac{\partial}{\partial a} (\frac{1}{2^{a}-1})$

we get $\Sigma_{n=1}^\infty -n2^{-an}\ln{2}=\frac{-2^{a}\ln{2}}{(2^{a}-1)^{2}}$

differentiate one more times $\Sigma_{n=1}^\infty -n^{2}2^{-an}\ln{2}=\frac{-2^{a}\ln{2}(2^{a}+1)}{(2^{a}-1)^{3}}$

rewriting the sum $\Sigma_{1}^\infty \frac{n(n+1)}{2}\frac{1}{2^{n}}=\Sigma_{1}^\infty\frac{1}{2}(\frac{n^{2}}{2^{n}}+\frac{n}{2^{n}})$

so we substitute a=1 in our sum,then we get $\frac{1}{2}(2+6)=\boxed{4}$