Sum of Sequence of Fifth Powers

Theorem

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

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

Basis for the Induction
$P \left({1}\right)$ is the case:

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

from which it is to be shown that:
 * $\displaystyle \sum_{i \mathop = 1}^{k + 1} i^5 = \dfrac { {T_{k + 1} }^2 \paren {4 T_{k + 1} - 1} } 3$

Induction Step
This is the induction step:

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

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