Prove by Mathematical Induction that the assertion,
∑ r^2 = n/6 (n+1)(2n+1)
holds for every natural number n.
Ok, so basically, how do you solve this question? I have got to the Induction step but I'm not sure how to do it.
The Attempt at a Solution
I've replaced n with k so I have,
1^2 + 2^2 + 3^2 + ... k^2 = k/6 (k+1)(2k+1)
Then I've added the (k+1)th term to each side to I have,
1^2 + 2^2 + 3^2 + ... k^2 + (k+1)^2 = k/6 (k+1)(2k+1) + (k+1)^2
So where do I go from here?