Sum of Sequences of Fifth and Seventh Powers

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Theorem

$\displaystyle \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$


Basis for the Induction

\(\displaystyle \sum_{i \mathop = 1}^1 i^5 + \sum_{i \mathop = 1}^1 i^7\) \(=\) \(\displaystyle 1^5 + 1^7\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2 \times 1^4\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {\sum_{i \mathop = 1}^1 i}^4\) $\quad$ $\quad$

So $\map P 1$ has been demonstrated to hold.

This is our basis for the induction.


Induction Hypothesis

Now we need to show 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 our induction hypothesis:

$\displaystyle \sum_{i \mathop = 1}^k i^5 + \sum_{i \mathop = 1}^k i^7 = 2 \paren {\sum_{i \mathop = 1}^k i}^4$


Then we need to show:

$\displaystyle \sum_{i \mathop = 1}^{k + 1} i^5 + \sum_{i \mathop = 1}^{k + 1} i^7 = 2 \paren {\sum_{i \mathop = 1}^{k + 1} i}^4$


Induction Step

This is our induction step:


\(\displaystyle 2 \paren {\sum_{i \mathop = 1}^{k + 1} i}^4\) \(=\) \(\displaystyle 2 \paren {\sum_{i \mathop = 1}^{k + 1} i^3}^2\) $\quad$ Sum of Sequence of Cubes $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {\paren {k + 1}^3 + \sum_{i \mathop = 1}^k i^3}^2\) $\quad$ Definition of Summation $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {k + 1}^6 + 4 \paren {\paren {k + 1}^3 \sum_{i \mathop = 1}^k i^3} + 2 \paren {\sum_{i \mathop = 1}^k i^3}^2\) $\quad$ Square of Sum $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {k + 1}^6 + 4 \paren {\paren {k + 1}^3 \frac {k^2 \paren {k + 1}^2} 4} + 2 \paren {\sum_{i \mathop = 1}^k i}^4\) $\quad$ Sum of Sequence of Cubes $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {k + 1}^6 + \paren {k^2 \paren {k + 1}^5} + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5\) $\quad$ Induction Hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {k + 1}^5 \paren {2 \paren {k + 1} + k^2} + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5\) $\quad$ simplifying first $2$ terms $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {k + 1}^5 \paren {\paren {k^2 + 2 k + 1} + 1} + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5\) $\quad$ simplifying $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {k + 1}^5 \paren {\paren {k + 1}^2 + 1} + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5\) $\quad$ Square of Sum $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {k + 1}^7 + \paren {k + 1}^5 + \sum_{i \mathop = 1}^k i^7 + \sum_{i \mathop = 1}^k i^5\) $\quad$ simplifying $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{k + 1} i^7 + \sum_{i \mathop = 1}^{k + 1} i^5\) $\quad$ Definition of Summation $\quad$

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

Therefore:

$\displaystyle \forall n \in \N_{>0}: \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$

$\blacksquare$


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