Prove that the sum of consecutive odd numbers beginning at 1 (eg 1, 3, 5, ..) always adds up to a perfect square

The ith term is given by T(i) = 2i-1
Let us denote the Sum to n terms as S(n).
Then,
S(n)
= Sum(i= 1 to n) (2i-1)
= 2*Sum(i= 1 to n) (i) - n
= n^2 + n -n
= n^2

Consequently, the required sum upto any term of the given sequence is always a perfect square.

Edited on September 11, 2023, 8:51 am

Posted by K Sengupta on 2007-03-24 12:00:10