Fourth Power as Summation of Groups of Consecutive Integers

Theorem
Take the positive integers and group them in sets such that the $m$th set contains the next $m$ positive integers:
 * $\set 1, \set {2, 3}, \set {4, 5, 6}, \set {7, 8, 9, 10}, \set {11, 12, 13, 14, 15}, \ldots$

Remove all the sets with an even number of elements.

Then the sum of all the integers in the first $n$ sets remaining equals $n^4$.

Proof
Let $S_m$ be the $m$th set of $m$ consecutive integers.

Let $S_k$ be the $k$th set of $m$ consecutive integers after the sets with an even number of elements have been removed.

Then $S_k = S_m$ where $m = 2 k - 1$.

By the method of construction:
 * the largest integer in $S_m$ is $T_m$, the $m$th triangular number
 * there are $m$ integers in $S_m$.

We also have that the middle integer in $S_m$ is $T_m - \dfrac {m - 1} 2$ (by inspection).

Thus the sum of the elements of $S_k$ is:

We need to calculate the sum of all $S_k$ from $1$ to $n$.

Hence we have: