If the sum of n terms of an A.P is cn (n-1)where `c ne 0` then the sum of the squares of these terms is

Options

A. `c^2n(n+1)^2`
B. `2/3c^2n(n-1)(2n-1)`
C. `(2c^2)/(3)n(n+1)(2n+1)`
D. none of these

Answer

B

Explanation

If `t_(r)` be the rth term of the A.P., then
`t_(r)=S_(r)-S_(r-1)`
`=cr(r-1)-c(r-1)(r-2)`
`=c(r-1)(r-r+2)-2c(r-1)`
We have,
`t_(1)^(2)+t_(2)^(2)+…+t_(n)^(2)=4c^(2)(0^(2)+1^(2)+2^(2)+…+(n-1)^(2))`
`=4c^(2)((n-1)n(2n-1))/6`
`=2/3c^2n(n-1)(2n-1)`