Using Mathematical Induction, prove the following
1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6
show that P(k) is true.
1^2 = 1(2)(3)/6 = 1
Assume P(k). Now,
1^2 + 2^2 + ... + k^2 + (k+1)^2 = k(k+1)(2k+1)/6 + (k+1)^2
= (k(k+1)(2k+1) + 6(k+1)^2)/6
= (k+1)(k(2k+1)+6(k+1))/6
= (k+1)(2k^2+k+6k+6)/6
= (k+1)(2k^2+7k+6)/6
= (k+1)(k+2)(2k+3)/6
= (k+1)(k+1)(2(k+1)+1)/6
= P(k+1)