Question

# If S1,S2,S3……,Sn,… are the sums of infinite geometric series whose first terms are 1,2,3,……n,…… and whose common ratios are 12,13,14,……1n+1…… respectively, then the value of 2n−1∑r=1S2r is

A
n(n+1)(2n+1)21
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B
n(n+1)(n+2)21
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C
n(2n+1)(4n+1)31
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D
n(2n+1)(4n+1)3(n+1)1
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Solution

## The correct option is C n(2n+1)(4n+1)3−1rth series will have a=r and common ratio is 1r+1 ∴Sr=r1−1r+1=r(r+1)r=r+1 ∴S2r=(r+1)2 ∴2n−1∑r=1S2r=2n−1∑r=1(r+1)2=22+32+42+……+(2n−1)2+(2n)2=12+22+32+……+(2n)2−1 = sum of the square of the first (2n) natural numbers =(2n)(2n+1)(4n+1)6−1=n(2n+1)(4n+1)3−1

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