Triangular Number as Alternating Sum and Difference of Squares

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Theorem

\(\ds \forall n \in \N: \ \ \) \(\ds \frac {n \paren {n + 1} } 2\) \(=\) \(\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j \paren {n - j}^2\)
\(\ds \) \(=\) \(\ds n^2 - \paren {n - 1}^2 + \paren {n - 2}^2 - \cdots + \paren {-1}^{n - 1}\)

Thus the $n$th triangular number can be expressed as the alternating sum and difference of squares:


So:

\(\ds 1\) \(=\) \(\ds 1^2\)
\(\ds 3\) \(=\) \(\ds 2^2 - 1^2\)
\(\ds 6\) \(=\) \(\ds 3^2 - 2^2 + 1^2\)
\(\ds 10\) \(=\) \(\ds 4^2 - 3^2 + 2^2 - 1^2\)

and so on.


Proof

The proof proceeds by induction.

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

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


Basis for the Induction

$\map P 1$ is the case:

\(\ds \frac {n \paren {n + 1} } 2\) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds 1^2\)

which 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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:


\(\ds \sum_{j \mathop = 0}^k \paren {-1}^j \paren {k - j + 1}^2\) \(=\) \(\ds \paren {k + 1}^2 - \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^j \paren {n - j}^2}\)
\(\ds \) \(=\) \(\ds \paren {k + 1}^2 - \frac {k \paren {k + 1} } 2\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \frac {2 k^2 + 4 k + 2 - k^2 - k} 2\) multiplying everything out
\(\ds \) \(=\) \(\ds \frac {k^2 + 3 k + 2} 2\) simplifying
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1} \paren {k + 2} } 2\) factoring


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

$\blacksquare$


Sources