# Sum of Sequences of Fifth and Seventh Powers

## Theorem

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

## Proof

Proof by induction:

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

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

### Basis for the Induction

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

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:

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

Then we need to show:

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

### Induction Step

This is our induction step:

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

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

Therefore:

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

$\blacksquare$