Find the sum of n terms of the series `1^2 + 4^2+ 7^2 + .........`

Explanation

` n^(th)` term of the given series ,` a_(n) = [1 + (n- 1) xx3]^(2)`
` = (3n- 2 )^(2) = 9n^(2) - 12n + 4 `
If ` S_(n)` be the sum to n terms of the given series , then
`S_(n) = sum_(n=1)^(n) a_(n)`
=` sum_(n=1)^(n) (9n^(2) - 12 + 4)`
= ` 9 sum _(n=1)^(n) n^(2) -12sum_(n=1)^(n) n+4n`
`= 9 (n(n+1)(2n +1))/(6) - (12-n(n+1))/(2) + 4n `
` = (3n(n+1)(2n+1))/(2) - (12n(n+1))/(2) + 4n`
`= n[(3n(n+1)(2n+1)-12(n+1)+8)/(2)]`
`= n[(3(2n^(2) + 3n +1)- 12(n+1) +8)/(2)]`
`= n ((6n^(2)+ 9n+ 3 - 12n - 12 + 8 )/(2))`
` = n(6n^(2) + 3n -1)/(2)`