Sum of Sequence of Seventh Powers

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Theorem

$\ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

$\ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$


$\map P 0$ is the case:

\(\ds \sum_{j \mathop = 0}^0 j^7\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds \frac {0^2 \paren {0 + 1}^2 \paren {3 \times 0^4 + 6 \times 0^3 - 0^2 - 4 \times 0 + 2} } {24}\)

Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

\(\ds \) \(\) \(\ds \frac {1^2 \paren {1 + 1}^2 \paren {3 \times 1^4 + 6 \times 1^3 - 1^2 - 4 \times 1 + 2} } {24}\)
\(\ds \) \(=\) \(\ds \frac {1 \times 2^2 \paren {3 + 6 - 1 - 4 + 2} } {24}\)
\(\ds \) \(=\) \(\ds \frac {4 \times 6} {24}\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds 1^7\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^1 j^7\)


Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds \sum_{j \mathop = 0}^k j^7 = \dfrac {k^2 \paren {k + 1}^2 \paren {3 k^4 + 6 k^3 - k^2 - 4 k + 2} } {24}$


from which it is to be shown that:

$\ds \sum_{j \mathop = 0}^{k + 1} j^7 = \dfrac {\paren {k + 1}^2 \paren{k + 2}^2 \paren {3 \paren {k + 1}^4 + 6 \paren {k + 1}^3 - \paren {k + 1}^2 - 4 \paren {k + 1} + 2} } {24}$


Induction Step

This is the induction step:


\(\ds \sum_{j \mathop = 0}^{k + 1} j^7\) \(=\) \(\ds \sum_{j \mathop = 0}^k j^7 + \paren {k + 1}^7\)
\(\ds \) \(=\) \(\ds \frac {k^2 \paren {k + 1}^2 \paren {3 k^4 + 6 k^3 - k^2 - 4 k + 2} } {24}\) Induction Hypothesis
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {k + 1}^2 \paren {k^5 + 5 k^4 + 10 k^3 + 10 k^2 + 5 k + 1}\) Binomial Theorem
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^2} {24} \paren {3 k^6 + 6 k^5 - k^4 - 4 k^3 + 2 k^2 + 24 k^5 + 120 k^4 + 240 k^3 + 240 k^2 + 120 k + 24}\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^2} {24} \paren {3 k^6 + 30 k^5 + 119 k^4 + 236 k^3 + 242 k^2 + 120 k + 24}\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^2 \paren {k + 2}^2} {24} \paren {3 k^4 + 18 k^3 + 35 k^2 + 24 k + 6}\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^2 \paren {k + 2}^2} {24} \paren {3 k^4 + 12 k^3 + 18 k^2 + 12 k + 3 + 6 k^3 + 18 k^2 + 18 k + 6 - k^2 - 2 k - 1 - 4 k - 4 + 2}\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}^2 \paren {k + 2}^2 \paren {3 \paren {k + 1}^4 + 6 \paren {k + 1}^3 - \paren {k + 1}^2 - 4 \paren {k + 1} + 2} } {24}\) Binomial Theorem

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


Therefore:

$\forall n \in \Z_{\ge 0}: \ds \sum_{j \mathop = 0}^n j^7 = \dfrac {n^2 \paren {n + 1}^2 \paren {3 n^4 + 6 n^3 - n^2 - 4 n + 2} } {24}$

$\blacksquare$


Also see


Sources

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