is this relation true? i found this relation in old notes , it's mentioned with no proof , and i want to know if it's true or false 
the relation says , 
$$ \sum_{k=1}^n n^6 =  \frac{n^7}{7} + \frac{n^6}{2} + \frac{5}{2} n^5 - \frac{15}{2}n^4 - \frac{31}{6}n^3 + \frac{11}{21}n - 8 $$
is this relation true or not ? 
and how can we prove it ? 
i think that using induction " if it is true " will be so so hard ! 
so , any other ideas ? 
 A: Try it with $n=1$. You get -17, which is clearly incorrect. I don't know how to derive an answer nicely, but you can generate eight points, then draw the 7th degree polynomial through them:
$$\frac{n^7}{7} + \frac{n^6}{2} + \frac{n^5}{2} - \frac{n^3}{6} + \frac{n}{42}$$
A: The relation is not correct. You want
$$\sum_{k=1}^n k^6=\frac{1}{7}n^7+\frac{1}{2}n^6+\frac{1}{2}n^5-\frac{1}{6}n^3+\frac{1}{42}n.$$
As with any of these identities, it is possible to verify the result by induction, once you have a correct conjecture. The calculation is mildly unpleasant, but utterly mechanical. 
There are various ways to derive the result, if we do not know it in advance.  
For a detailed discussion, see Faulhaber's Formula.
A: Use $$(r+1)^{n+1}-r^{n+1}=\sum_{0\le s\le n}\binom {n+1}s  r^s$$
to find $\sum_{1\le t \le n}r^s$  for $s=1,2,3,5,6$
For $s=0, \sum_{1\le r \le n}r^0=\sum_{1\le r \le n}1=n$
For $n=2, (r+1)^2-r^2=2r+1$
Put $r=1,2,\cdots,n-1,n$ and add to get $$(n+1)^2-1^2=2\sum_{1\le s\le n}r+\sum_{1\le s\le n}1$$
$$\implies 2\sum_{0\le s\le n}r=(n+1)^2-(n+1)\implies \sum_{0\le s\le n}r=\frac{n(n+1)}2$$
For $n=2+1=3,(r+1)^3-r^3=3r^2+3r+1$
Put $r=1,2,\cdots,n-1,n$ and add to get $$(n+1)^3-1^3$$
$$=3\sum_{1\le s\le n}r^2+3\sum_{1\le s\le n}r+\sum_{1\le s\le n}1$$
$$\implies 3\sum_{0\le s\le n}r^2=(n+1)^3-1-3\sum_{1\le s\le n}r-\sum_{1\le s\le n}1$$
$$=(n+1)^3-1-3\frac{n(n+1)}2-n=\frac{n(n+1)(2n+1)}2$$
and so on
