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.

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