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

Answer

C

Explanation

We have, `S_(n)=cn^(2)`
`:.` First term = c and common difference d=2c
So, given A.P. is c,3c,5c,7c, . . . . . .
Let S be the sum of first n terms of A.P. then,
`S=underset(r=1)overset(n)sum(2r-1)^(2)c^(2)`
`rArr" "S=c^(2)underset(r=1)overset(n)sum(4r^(2)-4r+1)`
`rArr" "S=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}=(n(4n^(2)-1))/(3)c^(2)`
ALITER We have, `S_(n)=cn^(2)`
`:." "a_(n)=S_(n)-S_(n-1)=cn^(2)-c(n-1)^(2)=(2n-1)c`
Let S be the sum of n terms of the sequence. Then,
`S=underset(r=1)overset(n)suma_(r)^(2)`
`rArr" "S=underset(r=1)overset(n)sum(2r-1)^(2)c^(2)`
`rArr" "S=c^(2)underset(r=1)overset(n)sum(4r^(2)-4r+1)`
`rArr" "S=c^(2){(4n(n+1)(2n+1))/(6)-(4n(n+1))/(2)+n}=(n(4n^(2)-1))/(3)c^(2)`