Anoher way:
(a)\;T:\mathbb{R}_5[x]\rightarrow{\mathbb{R}_5[x]} defined by T(p(x))=p(x+1)-p(x) is a linear map.
$(b)\;$ If $Y=AX$ is the equation of $T$ with respect to the canonical basis of $\mathbb{R}_5[x]$, using x^4\equiv(0,0,0,0,1,0)^t we get T^{-1}(x^4)\equiv (\alpha,-1/30,0,1/3,-1/2,1/5)^t with \alpha \in \mathbb{R} that is, $$T^{-1}(x^4)=\left\{{\alpha -x/30+x^3/3-x^4/2+x^5/5:\alpha \in{\mathbb{R}}}\right\}$$ $(c)$ Choose any polynomial h(x)\in{T^{-1}(x^4)} (for example, the corresponding to \alpha=0) . Such polynomial satisfies T(h(x))=x^4 i.e. h(x+1)-h(x)=x^4\;(*). For x=1,2,\ldots,n in (*) we get:
<br />
h(2)-h(1)=1^4\\<br />
h(3)-h(2)=2^4\\<br />
h(4)-h(3)=3^4\\<br />
\ldots\\<br />
h(n+1)-h(n)=n^4
That is, h(n+1)-h(n)=1^4+2^4+\ldots+n^4=S_4, hence
S_4=h(n+1)-h(1)=\\-\displaystyle\frac{n+1}{30}+\displaystyle\frac{(n+1)^3}{3}-\displaystyle\frac{(n+1)^4}{2}+\displaystyle\frac{(n+1)^5}{5}+\displaystyle\frac{1}{30}-\displaystyle\frac{1}{3}+\displaystyle\frac{1}{2}-\displaystyle\frac{1}{5}
Simplifying:
S_4=1^4+2^4+3^4+\ldots+n^4=\dfrac{n(2n+1)(n+1)(3n^2+3n-1)}{30}
P.S. More details here: http://www.fernandorevilla.es/docencia-problemas/problemas-2/2-s_414-n4-y-endomorfismo
