Even Power of 3 as Sum of Consecutive Positive Integers

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Theorem

Take the positive integers and group them in sets such that the $n$th set contains the next $3^n$ positive integers:

$\set 1, \set {2, 3, 4}, \set {5, 6, \ldots, 13}, \set {14, 15, \cdots, 40}, \ldots$

Let the $n$th such set be denoted $S_{n - 1}$, that is, letting $S_0 := \set 1$ be considered as the zeroth.


Then the sum of all the elements of $S_n$ is $3^{2 n}$.


Proof

The total number of elements in $S_0, S_1, \ldots, S_r$ is:

\(\ds \sum_{j \mathop = 0}^r \card {S_j}\) \(=\) \(\ds \sum_{j \mathop = 0}^r 3^j\)
\(\ds \) \(=\) \(\ds \dfrac {3^{r + 1} - 1} {3 - 1}\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds \dfrac {3^{r + 1} - 1} 2\) simplifying


Thus for any given $S_n$:

$\ds \sum S_n = \sum k \sqbrk {\dfrac {3^n - 1} 2 < k \le \dfrac {3^{n + 1} - 1} 2}$

using Iverson's convention.

So $\ds \sum S_n$ can be evaluated as the difference between two triangular numbers:

\(\ds \sum S_n\) \(=\) \(\ds \sum_{k \mathop = 1}^{\paren {3^{n + 1} - 1} / 2} k - \sum_{k \mathop = 1}^{\paren {3^n - 1} / 2} k\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\dfrac {3^{n + 1} - 1} 2 \paren {\dfrac {3^{n + 1} - 1} 2 + 1} - \dfrac {3^n - 1} 2 \paren {\dfrac {3^n - 1} 2 + 1} }\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\dfrac {3^{n + 1} - 1} 2 \paren {\dfrac {3^{n + 1} - 1 + 2} 2} - \dfrac {3^n - 1} 2 \paren {\dfrac {3^n - 1 + 2} 2} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 8 \paren {\paren {3^{n + 1} - 1} \paren {3^{n + 1} + 1} - \paren {3^n - 1} \paren {3^n + 1} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 8 \paren {\paren {3^{2 n + 2} - 1} - \paren {3^{2 n} - 1} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \dfrac 1 8 \paren {3^{2 n + 2} - 3^{2 n} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 8 \paren {3^{2 n} \paren {3^2 - 1} }\) extracting $3^{2 n}$ as a factor
\(\ds \) \(=\) \(\ds 3^{2 n}\) Oh look, that second factor magically equals $8$

Hence the result.

$\blacksquare$


Examples

\(\ds 3^0\) \(=\) \(\ds 1\)
\(\ds 3^2\) \(=\) \(\ds 2 + 3 + 4\)
\(\ds 3^4\) \(=\) \(\ds 5 + 6 + \cdots + 13\)
\(\ds 3^6\) \(=\) \(\ds 14 + 15 + \cdots + 40\)


Historical Note

In his Curious and Interesting Numbers of $1986$, David Wells attributes this result to M.N. Khatri, referencing its appearance in Volume $20$ of Scripta Mathematica, but it is proving difficult to corroborate this.


Sources

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