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: 1^{2}+2^{2}+3^{2}+….+n^{2}=^{1}⁄_{6}n(n+1)(2n+1).

**Solution**

*Part 1*: 1+3+5+…+(2n−1)=½(2n−1)2n − 2×½n(n−1)=n^{2}

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 = n^{2}.

Bottom of fraction

= sum of odd numbers to 4n−1 minus sum of odd numbers to 2n−1 = (2n)^{2}−n^{2} = 3n^{2}

Therefore fraction = n^{2}⁄ 3n^{2} = ^{1}⁄_{3} ▇

*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)=1^{2}+2^{2}+3^{2}+….+n^{2} + 2S, if S=the sum of different numbers multiplied pairwise which we want.

Therefore 2S=[½n(n+1)]^{2}−^{1}⁄_{6}n(n+1)(2n+1)

Expanding and simplifying, S=^{1}⁄_{24}n(n+1)(3n^{2}−n−2).