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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 2n1r=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)31
rth series will have a=r and common ratio is 1r+1
Sr=r11r+1=r(r+1)r=r+1 S2r=(r+1)2
2n1r=1S2r=2n1r=1(r+1)2=22+32+42++(2n1)2+(2n)2=12+22+32++(2n)21
= sum of the square of the first (2n) natural numbers =(2n)(2n+1)(4n+1)61=n(2n+1)(4n+1)31

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