Prove k^2=(1^2) +(2^2) + (3^2) +....+(n^2)= [n(n+1)(2n+1)/6] using induction.

Question

Prove k^2=(1^2) +(2^2) + (3^2) +....+(n^2)= [n(n+1)(2n+1)/6] using induction.

Answer 1

Prove that 1² + 2² + ... + n² = [n(n+1)(2n+1)]/6, for every positive integer n.

 

Proof:

 

I. Prove that the equation holds for n = 1

 

    If n = 1, then [n(n+1)(2n+1)]/6 = [(1)(2)(3)]/6 = 1.

       So, 1² = [(1)(1+1)(2(1)+1)]/6

 

II.  Assume that the equation holds for n and prove that          the equation holds for n+1.

 

       Assume:  1² + 2² + ... + n² = [n(n+1)(2n+1)]/6

 

        Prove:  1² + 2² + ... + n² +(n+1)² = 

                                                   [(n+1)(n+2)(2n+3)]/6

 

         By assumption, 1² + 2² +... + n² + (n+1)² = 

 

                                    [n(n+1)(2n+1)]/6 + (n+1)²

 

                                 = [n(n+1)(2n+1) +6(n+1)²]/6

 

                                 = [(n+1){n(2n+1) + 6(n+1)}]/6

 

                                 =[(n+1){2n² + 7n +6}]/6

 

                                 = [(n+1)(n+2)(2n+3)]/6

 

Therefore, by induction on n, the equation is valid for all positive integers, n.