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a
c2nn+12
b
23c2nn−12n−1
c
2c23nn+12n+1
d
none of these
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Correct option is B
If tr=Sr−Sr−1 =cr(r−1)−c(r−1)(r−2) =c(r−1)(r−r+2)=2c(r−1) We have, t12+t22+………+tn2=4c202+12+22+………+(n−1)2 =4c2(n−1)n(2n−1)6 =23c2n(n−1)(2n−1)Similar Questions
If the sum of first terms of an is , then the sum of squares of these terms is
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