<s>[INST] Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$ Find a polynomial expression for
$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$
I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?
 [/INST] Ignoring big theorems,
$$ f(n) = 1^4+2^4+\ldots+n^4 $$
is quite blatantly a polynomial in $n$ with degree $5$. It follows that it is enough to compute $f(0),f(1),\ldots,f(5)$ to recover the coefficients of such a polynomial by Lagrange interpolation:
$$ f(x) = \sum_{k=0}^{5}\,f(k)\!\!\!\!\prod_{\substack{j\in\{0,\ldots,5\}\\ j\neq k}}\frac{x-j}{k-j}.$$
Alternative approach: since
$$ \sum_{n\geq 0} n^4 e^{-nx} = \frac{d^4}{dx^4}\sum_{n\geq 0}e^{-nx} = \frac{d^4}{dx^4}\left(\frac{1}{1-e^{-x}}\right) =\frac{e^x(e^x+1)(e^{2x}+10 e^{x}+1)}{(e^x-1)^5}$$
it follows that for any $x\in(0,1)$ we have
$$ \sum_{n\geq 0}n^4 x^n = \frac{x \left(1+11 x+11 x^2+x^3\right)}{(1-x)^5} $$
and $f(N)$ is the coefficient of $x^N$ in $\frac{x \left(1+11 x+11 x^2+x^3\right)}{(1-x)^6}$. By stars and bars,
$$\boxed{\,f(N) = 1^4+2^4+\ldots+N^4= \color{red}{\frac{N(N+1)(2N+1)(3N^2+3N-1)}{30}} }$$
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