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If the sum of first n terms of an A P is...

If the sum of first `n` terms of an `A P` is `c n^2,` then the sum of squares of these `n` terms is (2009)

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
Let `S_(n) = cn^(2)`
`S_(n-1) = c(n-1)^(2) = cn^(2) + c - 2cn`
`:. T_(n) = 2cn -c " " [ :' T_(n) = S_(n) - S_(n-1)]`
`T_(n)^(2) = (2cn -c)^(2) = 4c^(2) n^(2) + c^(2) - 4c^(2) n`
`:.` Sum `= Sigma T_(n)^(2) = (4c^(2).n (n +1)(2n +1))/(6) + nc^(2) -2c^(2)n (n +1)`
`= (2c^(2) n (n +1) (2n +1) + 3nc^(2) - 6c^(2) n(n +1))/(3)`
`= (nc^(2) (4n^(2) + 6n + 2 + 3 - 6n -6))/(3) = (nc^(2) (4n^(2) -1))/(3)`
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