Question

Induction proof.

The series is the sum of squares = (n)(n+1)(2n+1)/6

Answer 1

\[\sum_{k=0}^{\inf} k^2 = \frac{ (n)(n+1)(2n+1) }{6 }\]

Answer 2

hmm it seems the next term would be (k+1)^2 so\[\sum_{0}^{\inf}(n^2)+a_{n+1} = \frac{ (n)(n+1)(2n+1) }{ 6 } + (n+1)^2 = \frac{ (n)(n+1)(2n+1)+ 6(n+1)^2 }{ 6 }\]

Answer 3

\[S_{n+1} = (n+1)\frac{ (n)(2n+1)+ 6(n+1) }{ 6 } = (n+1)\frac{ 2n^2 + n + 6n + 6 }{ 6 }\]

Answer 4

\[S_{n+1} = (n+1)\frac{ 2n^2 + 7n + 6 }{ 6 }\]
factor\[S_{n+1} = \frac{ (n+1)(n+2) (2n + 3) }{ 6 }\]
and done