If the sum of first n terms of an A.P. is `cn^(2)` then the sum of squares of these n terms is

Options

A. `(n(4n^(2)-1)c^(2))/(6)`
B. `(n(4n^(2)+1)c^(2))/(3)`
C. `(n(4n^(2)-1)c^(2))/(3)`
D. `(n(4n^(2)+1)c^(2))/(6)`

Answer

C

Explanation

`:.S_(n)=cn^(2)`
`:. T_(n)=S_(n)-S_(n-1)=c(2n-1)`
`sumt_(n)^(2)=c^(2)sum(2n-1)^(2)`
`=c^(2)sum(4n^(2)-4n+1)=c^(2){4sumn^(2)-4sumn+sum1}`
`=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}`
`=c^(2)n{(2)/(3)(2n^(2)3n+1)-2n-2+1}`
`=(c^(2)n)/(3)(4n^(2)-1)=(n(4n^(2)-1)c^(2))/(3)`.