Sums of Partial Sequences of Squares

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Theorem

Let $n \in \Z_{>0}$.

Consider the odd number $2 n + 1$ and its square $\paren {2 n + 1}^2 = 2 m + 1$.


Then:

$\ds \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$


That is:

the sum of the squares of the $n + 1$ integers up to $m$

equals:

the sum of the squares of the $n$ integers from $m + 1$ upwards.


Proof

First we express $m$ in terms of $n$:

\(\ds \paren {2 n + 1}^2\) \(=\) \(\ds 4 n^2 + 4 n + 1\)
\(\ds \) \(=\) \(\ds 2 \paren {2 n^2 + 2 n} + 1\)
\(\ds \leadsto \ \ \) \(\ds m\) \(=\) \(\ds 2 n^2 + 2 n\)


We have:

\(\ds \) \(\) \(\ds \sum_{j \mathop = 1}^n \paren {m + j}^2 - \sum_{j \mathop = 0}^n \paren {m - j}^2\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^n \paren {\paren {m + j}^2 - \paren {m - j}^2} - m^2\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^n 2 j \paren {2 m} - m^2\) Difference of Two Squares
\(\ds \) \(=\) \(\ds 4 m \times \frac {n \paren {n + 1} } 2 - m^2\) Closed Form for Triangular Numbers
\(\ds \) \(=\) \(\ds m \paren {2 n^2 + 2 n} - m^2\)
\(\ds \) \(=\) \(\ds m^2 - m^2\)
\(\ds \) \(=\) \(\ds 0\)

Hence the result.

$\blacksquare$


Examples

Example: $3^2 = 4 + 5$

\(\ds 3\) \(=\) \(\ds 1 + 2\)
\(\ds 3^2\) \(=\) \(\ds 4 + 5\)
\(\ds 3^2 + 4^2\) \(=\) \(\ds 5^2\)


Example: $5^2 = 12 + 13$

\(\ds 5\) \(=\) \(\ds 2 + 3\)
\(\ds 5^2\) \(=\) \(\ds 12 + 13\)
\(\ds 10^2 + 11^2 + 12^2\) \(=\) \(\ds 13^2+ 14^2\)


Example: $7^2 = 24 + 25$

\(\ds 7\) \(=\) \(\ds 3 + 4\)
\(\ds 7^2\) \(=\) \(\ds 24 + 25\)
\(\ds 21^2 + 22^2 + 23^2 + 24^2\) \(=\) \(\ds 25^2 + 26^2 + 27^2\)


Sources

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