Triangular Number as Alternating Sum and Difference of Squares

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Theorem

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

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


So:

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

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:

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

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:


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


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

$\blacksquare$


Sources