# Sum of Sequence of Seventh Powers

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## Contents

## 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:

\(\displaystyle \sum_{j \mathop = 0}^0 j^7\) | \(=\) | \(\displaystyle 0\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {0^2 \left({0 + 1}\right)^2 \left({3 \times 0^4 + 6 \times 0^3 - 0^2 - 4 \times 0 + 2}\right)} {24}\) |

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

### Basis for the Induction

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

\(\displaystyle \) | \(\) | \(\displaystyle \dfrac {1^2 \left({1 + 1}\right)^2 \left({3 \times 1^4 + 6 \times 1^3 - 1^2 - 4 \times 1 + 2}\right)} {24}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {1 \times 2^2 \left({3 + 6 - 1 - 4 + 2}\right)} {24}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {4 \times 6} {24}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {24} {24}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1^7\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop = 0}^1 j^7\) |

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:

\(\displaystyle \sum_{j \mathop = 0}^{k + 1} j^7\) | \(=\) | \(\displaystyle \sum_{j \mathop = 0}^k j^7 + \left({k + 1}\right)^7\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {k^2 \left({k + 1}\right)^2 \left({3 k^4 + 6 k^3 - k^2 - 4 k + 2}\right)} {24}\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle k^7 + 7 k^6 + 21 k^5 + 35 k^4 + 35 k^3 + 21 k^2 + 7 k + 1\) | Binomial Theorem | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\left({k + 1}\right)^2 \left({k + 2}\right)^2 \left({3 \left({k + 1}\right)^4 + 6 \left({k + 1}\right)^3 - \left({k + 1}\right)^2 - 4 \left({k + 1}\right) + 2}\right)} {24}\) | after algebraic manipulations |

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

Therefore:

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

$\blacksquare$

## Sources

- Weisstein, Eric W. "Power Sum." From
*MathWorld*--A Wolfram Web Resource. http://mathworld.wolfram.com/PowerSum.html