discrete math question

Question

prove that 1^2 + 2^2 +...+n^2 = (1/6) n (n+1)(2n+1) if n is an integer >= 1

Answer 1

Use induction.

Base case: Suppose n=1. Then 1^2 + 2^2 +...+n^2 = 1^2 = 1 = (1/6)x1x2x3 = (1/6) n (n+1)(2n+1)

Now suppose n>1 and

A. 1^2 + 2^2 +...+n^2 = (1/6) n (n+1)(2n+1) . <- A is our induction assumption

We want to prove that equation B below is true (gotten by replacing n by n+1:

B. 1^2 + 2^2 +...+n^2 + (n+1)^2 = (1/6) (n+1) ((n+1)+1)(2(n+1)+1)

By our induction assumption A, we can rewrite B as:

B': (1/6) n (n+1)(2n+1) + (n+1)^2 = (1/6) (n+1) ((n+1)+1)(2(n+1)+1)

Now expand both sides of B' to show that B' is true, so so is B.