Sum of Sequences of Fifth and Seventh Powers
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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:
- $\ds \forall n \in \N_{>0}: \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Exercise $18.10 \ \text{(d)}$