Hi Marie,
What I see is that for the nth term you add together n integers and then square the sum. The question is what n integers do you add? Look at the terms before the nth term. The first term uses 1 integer, the second term used 2 integers, the third term uses 3 integers, ..., the (n1)st term used n1 integers. Thus in total you have already used
1 + 2 + 3 + ... + (n1) integers.
For each n let
k(n) = 1 + 2 + ... + (n1) = n(n1)/2
The sum that makes up the nth term then starts at k(n) + 1 and hence the nth term is
Does this help?
Penny
