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

Answer

N/a

Explanation

`T_(n)={"nth term of 1, 4, 7,..."}^(2)`
`={1+(n+1).3}^(2)=(3n-2)^(2)`
`=9n^(2)-12n+4`
`rArr" "S_(n)=9Sigman^(2)-12Sigman+4n`
`=(9)/(6)n(n+1)(2n+1)-(12)/(2)n(n+1)+4n`
`=n[(3)/(2)(n+1)(2n+1)-6(n+1)+4]`
`=[3(n+1)(2n+1)-12(n+1)8]`
`=(n)/(2)(6n^(2)+9n+3-12n+8)`
`=(1)/(2)n(6n^(2)-3n-1).`