S[N}= SUM(2N-1) where N=1 to n.
2SUM(N)-n=2n(n+1)/2-n=n^2+n-n=n^2, so sum to n terms is just n^2.
[Sum of natural numbers 1, 2, 3, etc: 1+2+3+...+n=(1+n)+(2+n-1)+(3+n-2)+... Each pair adds up to n+1 and there are n/2 such pairs; hence sum of them = n(n+1)/2 whether n is even or odd.]