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If the sum of first n terms of an AP is ...

If the sum of first n terms of an AP is `cn^(2)`, then the sum of squares of these n terms is

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)`

Text Solution

Verified by Experts

The correct Answer is:
C
`:.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)`.
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