An algebra problem by Coby Tran

It's possible to note that:

$\large \displaystyle \sum_{k=1}^{\infty} x^k = \frac{x}{1-x}$ .

Differentiating with respect to $x$ and multiplying by $x$ one gets:

$\large \displaystyle \sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$ .

Doing it again:

$\large \displaystyle \color{#20A900} \sum_{k=1}^{\infty} k^2 x^k = \frac{x(1-x^2)}{(1-x)^4}$ .

So:

$\large \displaystyle 64 \sum_{k=1}^{\infty} k^2 \frac{3^{k+3}}{4^{k+4}}$ .

Becomes

$\large \displaystyle = 64\cdot \frac{27}{256} \sum_{k=1}^{\infty} k^2 \Big( \frac34 \Big)^k$ .

$\large \displaystyle = \frac{27}{4} \cdot \frac{(3/4)(1 - (3/4)^2)}{(1 - 3/4)^4}$

$\large \displaystyle = \frac{27}{4} \cdot \frac34 \cdot \frac{7}{16} \cdot 256$

$\large \displaystyle \color{#3D99F6} = \fbox{567}$