Sum of Sequence of Fifth Powers

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Theorem

$\ds \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$

where $T_n$ denotes the $n$th triangular number.


Proof

The proof proceeds by induction.

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

$\ds \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$


Basis for the Induction

$\map P 1$ is the case:

\(\ds \dfrac { {T_1}^2 \paren {4 T_1 - 1} } 3\) \(=\) \(\ds \dfrac {\paren {\frac {1 \paren {1 + 1} } 2}^2 \paren {4 \frac {1 \paren {1 + 1} } 2 - 1} } 3\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds \dfrac {1^2 \times \paren {4 \times 1 - 1} } 3\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^1 i^5\)

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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds \sum_{i \mathop = 1}^k i^5 = \dfrac { {T_k}^2 \paren {4 T_k - 1} } 3$


from which it is to be shown that:

$\ds \sum_{i \mathop = 1}^{k + 1} i^5 = \dfrac { {T_{k + 1} }^2 \paren {4 T_{k + 1} - 1} } 3$


Induction Step

This is the induction step:


\(\ds \sum_{i \mathop = 1}^{k + 1} i^5\) \(=\) \(\ds \sum_{i \mathop = 1}^k i^5 + \paren {k + 1}^5\)
\(\ds \) \(=\) \(\ds \dfrac { {T_k}^2 \paren {4 T_k - 1} } 3 + \paren {k + 1}^5\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \dfrac {\paren {\frac {k \paren {k + 1} } 2}^2 \paren {4 \frac {k \paren {k + 1} } 2 - 1} } 3 + \paren {k + 1}^5\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds \dfrac {\paren {k \paren {k + 1} }^2 \paren {2 k \paren {k + 1} - 1} } {12} + \paren {k + 1}^5\)
\(\ds \) \(=\) \(\ds \paren {k + 1}^2 \dfrac {\paren {2 k^3 \paren {k + 1} - k^2} + 12 \paren {k + 1}^3} {12}\)
\(\ds \) \(=\) \(\ds \paren {k + 1}^2 \dfrac {2 k^4 + 2 k^3 - k^2 + 12 k^3 + 36 k^2 + 36 k + 12} {12}\)
\(\ds \) \(=\) \(\ds \paren {k + 1}^2 \dfrac {2 k^4 + 14 k^3 + 35 k^2 + 36 k + 12} {12}\)
\(\ds \) \(=\) \(\ds \paren {k + 1}^2 \paren {k + 2}^2 \dfrac {2 k^2 + 6 k + 3} {12}\) extracting $\paren {k + 2}^2$ as a factor
\(\ds \) \(=\) \(\ds \paren {\frac {\paren {k + 1} \paren {k + 2} } 2}^2 \dfrac {4 \frac {\paren {k + 1} \paren {k + 2} } 2 - 1} 3\)
\(\ds \) \(=\) \(\ds \dfrac { {T_{k + 1} }^2 \paren {4 T_{k + 1} - 1} } 3\) Closed Form for Triangular Numbers

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


Therefore:

$\forall n \in \Z_{> 0}: \ds \sum_{i \mathop = 1}^n i^5 = \dfrac { {T_n}^2 \paren {4 T_n - 1} } 3$

$\blacksquare$


Also see


Sources

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