Can the following infinite sum be further simplified $\sum_{n=1}^{\infty}(n^2/4+n/4)1/5*(4/5)^{n-1}$?

Question

Can the following sum be further simplified? $${1\over 20}\sum_{n=1}^{\infty}\left(n^2+n\right)\left(\frac45\right)^{n-1}$$ (It's part of a probability problem)

Answer 1

$$f(x)=\sum_{n=1}^{\infty}n(n+1)x^{n-1}= (\sum_{n=1}^{\infty}(n+1)x^{n})'=(\sum_{n=1}^{\infty}x^{n+1})''$$

$$=\Big({x^2\over 1-x}\Big)''=\Big({2x-x^2\over (1-x)^2}\Big)'= {2\over (1-x)^3}$$

So $${1\over 20}f\Big({4\over 5}\Big) = 12,5$$

Answer 2

$$f(x)=\sum_{n=1}^{\infty}n(n+1)x^{n-1}=\frac 1x\sum_{n=1}^{\infty}[n(n-1)+2n]x^{n}$$ $$f(x)=\frac {x^2}x\sum_{n=1}^{\infty}n(n-1)x^{n-2}+\frac {2x}x\sum_{n=1}^{\infty}nx^{n-1}=x \left(\sum_{n=1}^{\infty}x^{n}\right)''+2\left(\sum_{n=1}^{\infty}x^{n}\right)'$$ Use now $$\sum_{n=1}^{\infty}x^{n}=\frac 1{1-x}$$