The sum of powers $\sum_{k=0}^n k^4$. I would like to ask if anyone can explain to me how can I find the closed form of the sum $$\sum_{k=0}^n k^4$$ with the expansion contraction method of the sum
Like this example : $$Y_n=\sum_{k=0}^n k^2$$
Yet  another  way  to  discover  a  closed  form  for  Cl,  is  to  replace the  original  sum  by  a  seemingly  more  complicated  double  sum  that  can  actually  be simplified  if  we  massage  it  properly 
$$\sum_{k=0}^n k^2 =\sum_{0\le j \le k \le n } k= \sum_{0\le j \le n} \sum k=\sum_{0\le j \le n}\frac{(j+n)(n-j+1)}{2} =\frac{1}{2}\sum_{0\le j \le n}(n(n+l)+j-j2)=  \frac{1}{2}n^2(n+1)+\frac{1}{4}n(n+1)-\frac{1}{2}Yn$$
 A: An approach via linear maps: 
1. $T:\mathbb{R}_5[x]\rightarrow{\mathbb{R}_5[x]},\quad T(p(x))=p(x+1)-p(x)$ is a linear map.
2.  The matricial equation of $T$ with respect to the canonical basis of $\mathbb{R}_5[x]$ is$$\begin{bmatrix}{y_1}\\{y_2}\\{y_3}\\{y_4}\\{y_5}\\{y_6}\end{bmatrix}=\begin{bmatrix}{0}&{1}&{1}&{1}&{1}&{1}\\{0}&{0}&{2}&{3}&{4}&{5}\\{0}&{0}&{0}&{3}&{6}&{10}\\ {0}&{0}&{0}&{0}&{4}&{10}\\{0}&{0}&{0}&{0}&{0}&{5}\\{0}&{0}&{0}&{0}&{0}&{0}\end{bmatrix}\begin{bmatrix}{x_1}\\{x_2}\\{x_3}\\{x_4}\\{x_5}\\{x_6}\end{bmatrix}\quad (*)$$
3. Using $(*)$ we get: $$T^{-1}(x^4)=\left\{{\alpha -x/30+x^3/3-x^4/2+x^5/5:\alpha \in{\mathbb{R}}}\right\}$$
4. Consider $h(x)\in{T^{-1}(x^4)}$ with $\alpha=0$. Then $T(h(x))=x^4$ i.e. $h(x+1)-h(x)=x^4$. 
5. For $x=1,2,\ldots,n$ we get:
$$\begin{aligned}
&h(2)-h(1)=1^4\\
&h(3)-h(2)=2^4\\
&h(4)-h(3)=3^4\\
&\ldots\\
&h(n+1)-h(n)=n^4.
\end{aligned}$$
6. Simplifying: $h(n+1)-h(n)=1^4+2^4+\ldots+n^4=S_4$. That is
$$S_4=h(n+1)-h(1)=$$ $$-\displaystyle\frac{n+1}{30}+\displaystyle\frac{(n+1)^3}{3}-\displaystyle\frac{(n+1)^4}{2}+\displaystyle\frac{(n+1)^5}{5}+\displaystyle\frac{1}{30}-\displaystyle\frac{1}{3}+\displaystyle\frac{1}{2}-\displaystyle\frac{1}{5}$$ $$=S_4=1^4+2^4+3^4+\ldots+n^4=\dfrac{n(2n+1)(n+1)(3n^2+3n-1)}{30}.$$
A: We have $$
(k+1)^5-k^5 = 5 k^4 + 10 k^3 + 10 k^2 + 5 k + 1
$$
and so
$$
(n+1)^5=
\sum_{k=0}^n (k+1)^5-\sum_{k=0}^nk^5 =\\ = \sum_{k=0}^n 5 k^4 + 10 k^3 + 10 k^2 + 5 k + 1 = 5 S_4 + 10 S_3 + 10 S_2 + 5 S_1 + S_0
$$
If you know the formulas for $S_3, S_2, S_1, S_0$, then you can solve for $S_4$.
A: As $S_4(n)-S_4(n-1)=n^4$ is a quartic polynomial in $n$, $S_4(n)$ must be quintic.
You can obtain it as the Lagrangian interpolator of the six points 
$$(0,0),(1,1),(2,17),(3,98),(4,354),(5,979).$$
The computation is quite laborious, but gives the correct
$$S_4(n)=\frac{n^5}5+\frac{n^4}2+\frac{n^3}3-\frac n{30}.$$
A: Let $S_d(n):=\sum_{k=1}^n k^d$.
Then as $$k^d-(k-1)^d=dk^{d-1}-\binom d2k^{d-2}+\binom d3k^{d-3}-\cdots(-1)^d$$
we have
$$S_1(n)-S_1(n-1)=S_0(n)=n,$$
$$S_2(n)-S_2(n-1)=2S_1(n)-S_0=n^2,$$
$$S_3(n)-S_3(n-1)=3S_2(n)-3S_1(n)+S_0=n^3,$$
$$S_4(n)-S_4(n-1)=4S_3(n)-6S_2(n)+4S_1-S_0=n^4,$$
$$S_5(n)-S_5(n-1)=5S_4(n)-10S_3(n)+10S_2-5S_1+S_0=n^5.$$
Solving, we obtain
$$S_1(n)=\frac{n^2}2+\frac n2,$$
$$S_2(n)=\frac{n^3}3+\frac{n^2}2+\frac n6,$$
$$S_3(n)=\frac{n^4}4+\frac{n^3}2+\frac{n^2}4,$$
$$S_4(n)=\frac{n^5}5+\frac{n^4}2+\frac{n^3}3-\frac n{30}.$$
