Sum of Sequence of Seventh Powers

Theorem

 * $\displaystyle \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \left({n + 1}\right)^2 \left({3 n^4 + 6 n^3 - n^2 - 4 n + 2}\right)} {24}$

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \left({n + 1}\right)^2 \left({3 n^4 + 6 n^3 - n^2 - 4 n + 2}\right)} {24}$

$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_{j \mathop = 0}^k j^7 = \dfrac {k^2 \left({k + 1}\right)^2 \left({3 k^4 + 6 k^3 - k^2 - 4 k + 2}\right)} {24}$

from which it is to be shown that:
 * $\displaystyle \sum_{j \mathop = 0}^{k + 1} j^7 = \dfrac {\left({k + 1}\right)^2 \left({k + 2}\right)^2 \left({3 \left({k + 1}\right)^4 + 6 \left({k + 1}\right)^3 - \left({k + 1}\right)^2 - 4 \left({k + 1}\right) + 2}\right)} {24}$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: \displaystyle \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \left({n + 1}\right)^2 \left({3 n^4 + 6 n^3 - n^2 - 4 n + 2}\right)} {24}$