Question

1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

how do you solve for Sk+1

Answer 1

are you proving this by induction?

Answer 2

\[\sum_{i=1}^{n}(3n-2)^2=\frac{n(6n^2-3n-1)}{2}\]
this is true for n = 1
\[\sum_{i=1}^{n+1}(3n-2)^2=(3(n+1)-2)^2+\sum_{i=1}^{n}(3n-2)^2=(3(n+1)-2)^2+\frac{n(6n^2-3n-1)}{2}\]
\[=(3n+1)^2+\frac{n(6n^2-3n-1)}{2}=\frac{2(3n+1)^2+n(6n^2-3n-1)}{2}=\\\frac{2(9n^2+6n+1)+6n^3-3n^2-n}{2}=\frac{18n^2+12n+2+6n^3-3n^2-n}{2}=\\\frac{(n+1)(6(n+1)^2-3(n+1)-1)}{2}\]
\[\text{thus by induction}\]
\[\sum_{i=1}^{n}(3n-2)^2=\frac{n(6n^2-3n-1)}{2}\]