How Can I prove $\sum_{n=1}^{\infty}{\frac{n^4}{5^n}}=\frac{285}{128}$ Question:Prove that $ \sum_{n=1}^{\infty}{\frac{n^4}{5^n}}=\frac{285}{128}$
While doing questions on series and products,i got stuck in this question.I was not able to figure out any way how to prove this one.Convergence tests shows that this series converges but i don't know how to find its value.

Equivalently, we want to evaluate $\left.\sum_nn^4x^n\right|_{x=1/5}$ with $\sum_n:=\sum_{n=\color{blue}{0}}^\infty$. By the binomial theorem, if $|x|<1$ then$$\begin{align}(1-x)^{-1}&=\sum_nx^n,\,\\(1-x)^{-2}&=\sum_n(n+1)x^n,\,\\(1-x)^{-3}&=\sum_n\tfrac12(n+1)(n+2)x^n,\,\\(1-x)^{-4}&=\sum_n\tfrac16(n+1)(n+2)(n+3)x^n,\,\\(1-x)^{-5}&=\sum_n\tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)x^n.\end{align}$$We'll find a linear combination by first getting the $n^4$ coefficient right, then $n^3$ etc. Since$$\begin{align}n^4&=24\cdot\tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)\\&-60\cdot\tfrac16(n+1)(n+2)(n+3)\\&+50\cdot\tfrac12(n+1)(n+2)\\&-15\cdot(n+1)\\&+1,\end{align}$$we have$$\begin{align}\sum_nn^4x^n&=24(1-x)^{-5}-60(1-x)^{-4}\\&+50(1-x)^{-3}-15(1-x)^{-2}+(1-x)^{-1}.\end{align}$$Substituting $x=\tfrac15$ gives the desired result.
