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:

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


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

The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:

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

where $\paren {2 n + 1})^2 = 2 m + 1$.


First it is worth rewriting this so as to eliminate $m$.

\(\displaystyle \paren {2 n + 1}^2\) \(=\) \(\displaystyle 4 n^2 + 4 n + 1\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {2 n^2 + 2 n} + 1\)

Thus the statement to be proved can be expressed:

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


Basis for the Induction

$\map P 1$ is the case:

$3^2 + 4^2 = 5^2$

which is the $3-4-5$ triangle.


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:

$\displaystyle \sum_{j \mathop = 0}^k \paren {2 k^2 + 2 k - j}^2 = \sum_{j \mathop = 1}^k \paren {2 k^2 + 2 k + j}^2$


from which it is to be shown that:

$\displaystyle \sum_{j \mathop = 0}^{k + 1} \paren {2 \paren {k + 1}^2 + 2 \paren {k + 1} - j}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 \paren {k + 1}^2 + 2 \paren {k + 1} + j}^2$


Induction Step

This is the induction step:


\(\displaystyle \) \(\) \(\displaystyle \sum_{j \mathop = 0}^{k + 1} \paren {2 \paren {k + 1}^2 + 2 \paren {k + 1} - j}^2\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 0}^{k + 1} \paren {2 k^2 + 6 k + 4 - j}^2\)



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


Therefore:

$\displaystyle \forall n \in \Z_{>0}: \sum_{j \mathop = 0}^n \paren {m - j}^2 = \sum_{j \mathop = 1}^n \paren {m + j}^2$

where $\paren {2 n + 1}^2 = 2 m + 1$.

$\blacksquare$


Examples

Example: $3^2 = 4 + 5$

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


Example: $5^2 = 12 + 13$

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


Example: $7^2 = 24 + 25$

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


Sources