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Question

If the sum of first n terms of an AP is cn2, then the sum of squares of these n terms is

A
n (4n21)c26
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B
n (4n2+1)c23
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C
n (4n21)c23
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D
n (4n2+1)c26
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Solution

The correct option is C n (4n21)c23

Sn=cn2
Sn1=c(n1)2=cn2+c2cn
Tn=2cnc
T2n=(2cnc)2=4c2n2+c24c2n
Required sum
=T2n=4c2n2+nc24c2n
=4c2n(n+1)(2n+1)6+nc22c2n(n+1)
=2c2n(n+1)(2n+1)+3nc26c2n(n+1)3
=nc2[4n2+6n+2+36n6]3
=nc2(4n21)3


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