Fortnightly maths prize for 23 June

No-one attempted this one, which I guess is due to it being post-exam time rather than it being specially difficult.

Part 1:


To help you: 1+2+3+4+….+n=½n(n+1) for all n

Part 2:

Find the sum to n terms of the first n numbers multiplied pairwise

For n = 3 it is 1×2 + 1×3 + 2×3, which makes 11

Your job is to find a general formula which works for any n.

To help you: 12+22+32+….+n2=16n(n+1)(2n+1).


Part 1: 1+3+5+…+(2n−1)=½(2n−1)2n − 2×½n(n−1)=n2

Or, better, see this result directly from


So if the fraction goes up to (4n−1) on the bottom, the top

= sum of odd numbers to 2n−1 = n2.

Bottom of fraction

= sum of odd numbers to 4n−1 minus sum of odd numbers to 2n−1 = (2n)2−n2 = 3n2

Therefore fraction = n2⁄ 3n2 = 13   ▇

Part 2: (1+2+3)×(1+2+3) = 1×1+2×2+3×3 + 2(1×2+1×3+2×3)

Generally (1+2+3+4+….+n)×(1+2+3+4+….+n)=12+22+32+….+n2 + 2S, if S=the sum of different numbers multiplied pairwise which we want.

Therefore 2S=[½n(n+1)]216n(n+1)(2n+1)

Expanding and simplifying, S=124n(n+1)(3n2−n−2).