Questions

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

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n4n2−1c26

n4n2+1c23

n4n2−1c23

n4n2+1c26

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detailed solution

Correct option is C

∵Sn=cn2∴ tn=Sn−Sn−1=c(2n−1) Σtn2=c2Σ(2n−1)2=c2Σ4n2−4n+1=c24Σn2−4Σn+Σ1=c24n(n+1)(2n+1)6−4n(n+1)2+n=c2n232n2+3n+1−2n−2+1=c2n34n2−1=n4n2−1c23

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