Sum of Sequence of Seventh Powers

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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:

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

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


Basis for the Induction

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

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


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:


\(\displaystyle \sum_{j \mathop = 0}^{k + 1} j^7\) \(=\) \(\displaystyle \sum_{j \mathop = 0}^k j^7 + \left({k + 1}\right)^7\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {k^2 \left({k + 1}\right)^2 \left({3 k^4 + 6 k^3 - k^2 - 4 k + 2}\right)} {24}\) $\quad$ Induction Hypothesis $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle + \, \) \(\displaystyle k^7 + 7 k^6 + 21 k^5 + 35 k^4 + 35 k^3 + 21 k^2 + 7 k + 1\) $\quad$ Binomial Theorem $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\left({k + 1}\right)^2 \left({k + 2}\right)^2 \left({3 \left({k + 1}\right)^4 + 6 \left({k + 1}\right)^3 - \left({k + 1}\right)^2 - 4 \left({k + 1}\right) + 2}\right)} {24}\) $\quad$ after algebraic manipulations $\quad$



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


Therefore:

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

$\blacksquare$


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