Failure I

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

Differentiating both sides, we get that $\displaystyle \sum_{n=1}^{\infty} nx^{n-1} = \dfrac{1}{(1-x)^2} \Leftrightarrow \displaystyle \sum_{n=1}^{\infty} nx^n = \dfrac{x}{(1-x)^2}$ .

Differentiating both sides again, we finally get $S = \displaystyle \sum_{n=1}^{\infty} n^2x^{n-1} = \dfrac{x+1}{(1-x)^3} \Leftrightarrow \displaystyle \sum_{n=1}^{\infty} n^2x^n = \dfrac{x(x+1)}{(1-x)^3}$ .

Plugging $x = \dfrac{3}{4}$ , get $\boxed{S = 84.}$