# Useless Numerology for 2016: Part 5

The following entertaining (but useless) facts about the number 2,016 appeared in a recent Facebook post (and subsequent comments) by the American Mathematical Monthly.

$2016 = \displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2$

$2016 = 2^{11} - 2^5$

Establishing that these two expressions are equal takes a little bit of work. We begin by dividing the sum into even and odd values of $n$. The odd values of $n$ from $n = 1$ to $n = 63$ have the form $n = 2k+1$, where $k$ varies between $k = 0$ and $k = 31$. The even values of $n$ from $n = 0$ to $n = 62$ have the form $n = 2k$, where again $k$ varies between $k = 0$ and $k = 31$.

$\displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2 = \displaystyle \sum_{n=0 \atop n =2k+1}^{63} (-1)^{n+1} n^2 + \displaystyle \sum_{n=0 \atop n =2k}^{63} (-1)^{n+1} n^2$

$= \displaystyle \sum_{k=0}^{31} (-1)^{(2k+1)+1} (2k+1)^2 + \displaystyle \sum_{k=0}^{31} (-1)^{2k+1} (2k)^2$

$= \displaystyle \sum_{k=0}^{31} \left[ (-1)^{2k+2} (2k+1)^2 + (-1)^{2k+1} (2k)^2 \right]$

$= \displaystyle \sum_{k=0}^{31} \left[(2k+1)^2 - (2k)^2 \right]$

$= \displaystyle \sum_{k=0}^{31} \left[4k^2 + 4k + 1 - 4k^2 \right]$

$= \displaystyle \sum_{k=0}^{31} \left[4k + 1 \right]$

This last sum is an arithmetic series. The first ($k = 0$) term is $1$, the last term is $4(31) + 1$, and there are $32$ terms. Using the formula for an arithmetic series, we find

$\displaystyle \sum_{n=0}^{63} (-1)^{n+1} n^2= \displaystyle \frac{32 [1 + 4(31) + 1]}{2}$

$= \displaystyle \frac{32[4(31) + 2]}{2}$

$= 32[2(31) + 1]$

$= 32[2(32-1) + 1]$

$= 32[2(32) - 2 + 1]$

$= 32[64 - 1]$

$= 2^5[2^6 -1]$

$= 2^{11} - 2^5$.