Fourth Power as Summation of Groups of Consecutive Integers

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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:

\(\ds \sum S_k\) \(=\) \(\ds m \times \paren {T_m - \dfrac {m - 1} 2}\)
\(\ds \) \(=\) \(\ds m \times \paren {\dfrac {m \paren {m + 1} } 2 - \dfrac {m - 1} 2}\)
\(\ds \) \(=\) \(\ds m \times \paren {\dfrac {m \paren {m + 1} } 2 - \dfrac {m - 1} 2}\)
\(\ds \) \(=\) \(\ds m \times \paren {\dfrac {m^2 + 1} 2}\)
\(\ds \) \(=\) \(\ds \dfrac {m^3 + m} 2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {2 k - 1}^3 + \paren {2 k - 1} } 2\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {8 k^3 - 12 k^2 + 6 k - 1} + \paren {2 k - 1} } 2\)
\(\ds \) \(=\) \(\ds 4 k^3 - 6 k^2 + 4 k - 1\)


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


Hence we have:

\(\ds \sum_{k \mathop = 1}^n S_k\) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {4 k^3 - 6 k^2 + 4 k - 1}\)
\(\ds \) \(=\) \(\ds 4 \sum_{k \mathop = 1}^n k^3 - 6 \sum_{k \mathop = 1}^n k^2 + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1\)
\(\ds \) \(=\) \(\ds 4 \paren {\frac {n^2 \paren {n + 1}^2} 4} - 6 \sum_{k \mathop = 1}^n k^2 + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1\) Sum of Sequence of Cubes
\(\ds \) \(=\) \(\ds n^2 \paren {n + 1}^2 - 6 \paren {\frac {n \paren {n + 1} \paren {2 n + 1} } 6} + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1\) Sum of Sequence of Squares
\(\ds \) \(=\) \(\ds n^2 \paren {n + 1}^2 - n \paren {n + 1} \paren {2 n + 1} + 4 \paren {\frac {n \paren {n + 1} } 2} - \sum_{k \mathop = 1}^n 1\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds n^2 \paren {n + 1}^2 - n \paren {n + 1} \paren {2 n + 1} + 2 n \paren {n + 1} - n\) simplification
\(\ds \) \(=\) \(\ds n^4 + 2 n^3 + n^2 - 2 n^3 - 3 n^2 - n + 2 n^2 + 2 n - n\) multiplying everything out
\(\ds \) \(=\) \(\ds n^4\) simplifying

$\blacksquare$


Examples

Examples: $3^4$

$1 + \paren {4 + 5 + 6} + \paren {11 + 12 + 13 + 14 + 15} = 3^4$


Sources

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