Can you find this infinite sum?

Consider the summation:

$\begin{aligned} S & = \sum_{k=1}^\infty \frac {(-1)^{k+1}k^2}{5^{k-1}} \\ & = 1 - \frac 45 + \frac 9{5^2} - \frac {16}{5^3} + \frac {25}{5^4} - \cdots \\ \frac S5 & = \frac 15 - \frac 4{5^2} + \frac 9{5^3} - \frac {16}{5^4} + \frac {25}{5^5} - \cdots \\ S + \frac S5 & = 1 - \frac 35 + \frac 5{5^2} - \frac 7{5^3} + \frac 9{5^4} - \cdots \\ \frac 15 \left(S + \frac S5\right) & = \frac 15 - \frac 3{5^2} + \frac 5{5^3} - \frac 7{5^4} + \frac 9{5^5} - \cdots \\ \left(S + \frac S5\right) + \frac 15 \left(S + \frac S5\right) & =1 - \frac 25 + \frac 2{5^2} - \frac 2{5^3} + \frac 2{5^4} - \cdots \\ \implies \frac {36}{25}S & = 1 - \frac 25 \left(\frac 1{1+\frac 15}\right) = \frac 23 \\ S & = \frac {25}{54} \\ \sum_{k=1}^\infty \frac {54(-1)^{k+1}k^2}{5^{k-1}} & = 54 S = \boxed{25} \end{aligned}$

Consider the infinite series $\displaystyle \sum_{x=1}^\infty a^x = \dfrac a{1-a}$ for $|a| < 1$ .

Differentiating with respect to $a$ gives

$\displaystyle \sum_{x=1}^\infty x \cdot a^{x-1} = \dfrac 1{(1-a)^2}.$

Multiplying it by $a$ , we have

$\displaystyle \sum_{x=1}^\infty x \cdot a^x = \dfrac a{(1-a)^2}.$

Differentiating with respect to $a$ again gives

$\displaystyle \sum_{x=1}^\infty x^2 \cdot a^{x-1} = \dfrac {a+1}{(1-a)^3}.$

Multiplying it by $a$ , we have

$\displaystyle \sum_{x=1}^\infty x^2 \cdot a^{x-1} = \dfrac {a(a+1)}{(1-a)^3}.$

The series in question is equal to $(54\cdot 5 \cdot -1)\displaystyle \sum_{x=1}^\infty x^2 a^x$ , where $a = -\frac15$ . Upon substitution, we have the answer of $\boxed{25}$ .