If the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is
A
n(4n2−1)c26
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B
n(4n2+1)c23
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C
n(4n2−1)c23
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D
n(4n2+1)c26
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Solution
The correct option is Cn(4n2−1)c23 tn=Sn−Sn−1=c{n2−(n−1)2}=c(2n−1) ⇒t2n=c2(4n2−4n+1) ∑nn=1t2n=c2{4n(n+1)(2n+1)6−4n(n+1)2+n} =c2n6{4(n+1)(2n+1)−12(n+1)+6} =c2n3{4n2+6n+2−6n−6+3}=c23n(4n2−1)