## Alternating Sign Square Sum

#### Problem

It can be seen that $6^2 - 5^2 + 4^2 - 3^2 + 2^2 - 1^2 = 21 = T_6$, the sixth triangle number.

Prove that the $n$th triangle number, $T_n = n^2 - (n - 1)^2 + (n - 2)^2 - ... 1^2$.

#### Solution

The difference between two consecutive squares, $k^2 - (k - 1)^2 = 2k - 1$.

Let us consider even $n$, and group them into $n/2$ pairs:

\begin{align}S &= (n^2 - (n - 1)^2) + ... + (6^2 - 5^2) + (4^2 - 3^2) + (2^2 - 1^2)\\&= 2n - 1 + ... + 11 + 7 + 3\\&= 3 + ... + 2n-5 + 2n-1\\\therefore 2S &= \dfrac{n}{2}(2n + 2)\\&= n(n + 1)\\\therefore S &= \dfrac{n(n + 1)}{2}\end{align}

When $n$ is odd we have $(n - 1)/2$ pairs plus one odd term:

\begin{align}S &= (n^2 - (n - 1)^2) + ... + (5^2 - 4^2) + (3^2 - 2^2) + 1^2\\&= (2n - 1 + ... + 9 + 5) + 1\\\therefore 2S &= \dfrac{n-1}{2}(2n + 4) + 2\\&= (n - 1)(n + 2) + 2\\&= n^2 + n\\&= n(n + 1)\\\therefore S &= \dfrac{n(n + 1)}{2}\end{align}
Problem ID: 262 (29 Jan 2006)     Difficulty: 3 Star

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