Question 474337
Use induction. The base case n = 1 works. For n >= 2, if


*[tex \LARGE \sum_{i=1}^{n} (2n - 1) = n^2] holds, then the n+1 case works, i.e.


*[tex \LARGE \sum_{i=1}^{n+1} (2n - 1) = \sum_{i=1}^n (2n - 1) = n^2 + (2(n+1) - 1) = n^2 + 2n + 1 = (n+1)^2]


Hence, if the identity holds for n, it will hold for n+1 and we are done.