Now that's what i call a series!

$\sum_{n=0}^{ \infty } x^{n} = \dfrac{1}{1-x} \quad \forall \quad |x| \leq 1$ $\Rightarrow \sum_{n=0}^{ \infty } ( \frac{d}{dx} x^{n} ) = \frac{d}{dx} ( \dfrac{1}{1-x} ) \\ \Rightarrow \sum_{n=0}^{ \infty} ( n \cdot x^{n} ) = \dfrac{x}{ (1-x)^{2} } \\ \Rightarrow \sum_{n=0}^{ \infty} \frac{d}{dx} (n \cdot x^{n} ) = \frac{d}{dx} ( \dfrac{x}{ (1-x)^{2} } ) \\ \Rightarrow \sum_{n=0}^{ \infty} ( n^{2} \cdot x^{n} ) = x\cdot \dfrac{ (1-x)^{2} + 2x\cdot (1-x) }{ (1-x)^{4} } \\= \frac{x+x^{2}}{ (1-x)^{3} }$

Well, we just input $x=\frac{1}{4}$ and we get the answer as $\dfrac{20}{27}$ .

So $M + N = 20 + 27 = 47$

$\begin{aligned} \sum_{k=0}^\infty {\frac{k^2}{4^k}} & = \sum_{\color{#D61F06}{k=1}}^\infty {\frac{k^2}{4^k}} \\ & = \sum_{\color{#3D99F6}{k=0}}^\infty {\frac{(k+1)^2}{4^{k+1}}} \\ & = \frac{1}{4} \sum_{k=0}^\infty {\frac{k^2+2k+1}{4^k}} \\ \Rightarrow 4 \sum_{k=0}^\infty {\frac{k^2}{4^k}} & = \sum_{k=0}^\infty {\frac{k^2}{4^k}} + \sum_{k=0}^\infty {\frac{2k}{4^k}} + \sum_{k=0}^\infty {\frac{1}{4^k}} \\ \Rightarrow 3 \sum_{k=0}^\infty {\frac{k^2}{4^k}} & = 2 \sum_{k=0}^\infty {\frac{k}{4^k}} + \sum_{k=0}^\infty {\frac{1}{4^k}} \end{aligned}$

Similarly,

$\begin{aligned} \sum_{k=0}^\infty {\frac{k}{4^k}} & = \sum_{k=1}^\infty {\frac{k}{4^k}} = \sum_{k=0}^\infty {\frac{k+1}{4^{k+1}}} = \frac{1}{4} \sum_{k=0}^\infty {\frac{k}{4^k}} + \frac{1}{4} \sum_{k=0}^\infty {\frac{1}{4^k}} \\ \Rightarrow \frac{3}{4} \sum_{k=0}^\infty {\frac{k}{4^k}} & = \frac{1}{4} \sum_{k=0}^\infty {\frac{1}{4^k}} = \frac{1}{4} \left(\frac{1}{1-\frac{1}{4}} \right) = \frac{1}{3} \\ \Rightarrow \sum_{k=0}^\infty {\frac{k}{4^k}} & = \frac{4}{9} \end{aligned}$

Now, we have:

$\begin{aligned} \Rightarrow 3 \sum_{k=0}^\infty {\frac{k^2}{4^k}} & = 2 \sum_{k=0}^\infty {\frac{k}{4^k}} + \sum_{k=0}^\infty {\frac{1}{4^k}} = 2 \left(\frac{4}{9} \right) + \frac{4}{3} = \frac{20}{9} \\ \Rightarrow \sum_{k=0}^\infty {\frac{k^2}{4^k}} & = \frac{20}{27} \end{aligned}$

$\Rightarrow M + N = 20 + 27 = \boxed{47}$

S =1/4 +4/16+9/64 ....... S/4. = 1/16 +4/64........ 3S/4 = 1/4 +3 /16+ 5/64.....i 3S/16= 1/16+3/64......ii (ii)-(i) implies 9S/16= 1/4+ 2(1/16+1/64....) ( infinte G.P). It gives us S= 20/27...pretty simple but a useful method i guess...i always liked manipulation....others do have a nice approach too

You can also make recurrent sequence let $a_n$ be sum of first $n+1$ terms (include k=0).

$a_n=a_{n-1}+\frac{n^2}{4^n}$

Now the solution is:

$a_n=a_n'+a_n''$

$a_n'=C$

$a_n''=\frac{An^2+Bn+C}{4^n}$

Solution for $A, B, C$ are respectively $-\frac{1}{3}, -\frac{8}{9}, -\frac{20}{27}$ .

Putting $a_0=C+a_0''=0$ you will find

$C=\frac{20}{27}$

And now the final expression for $a_n$ :

$a_n=\frac{20}{27}-\frac{\frac{1}{3}n^2+\frac{8}{9}n+\frac{20}{27}}{4^n}$

Now letting $n$ goes to infinity second term goes to zero so the answer is

$\frac{20}{27}$

$20+27=47$