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Question

# Let Sn,n=1,2,3,........., be the sum of infinite geometric series whose first term is n and the common ratio is 1n+1. Evaluatelimn→∞S1Sn+S2Sn−1+S3Sn−2+...+SnS1S21+S22+.....+S2n

A
2
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B
4
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C
12
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D
14
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Solution

## The correct option is D 12Sn=n1−1n+1Sn=n+1∴S1Sn+S2Sn−1+S3Sn−2+...+SnS1=2.(n+1)+3.n+4.(n−1)+...+(n+1).2=∑nr=1(r+1)(n−r+2)=∑nr=1(nr−r2+2r+n−r+2)=∑nr=1{(n+1)r−r2+(n+2)}=(n+1)∑nr=1r−∑nr=1r2+(n+2)∑nr=11=(n+1)∑n−∑n2+(n+2).n=(n+1)n(n+1)2−n(n+1)(2n+1)6+n(n+2)=n6{3(n2+2n+1)−(2n2+3n+1)+6n+12}=n6{n2+9n+14}⋯(1)and S21+S22+.....+S2n=22+32+...+(n+1)2=(n+1)(n+2)(2n+3)6−1=n(2n2+9n+13)6⋯(2)from (1) and (2), we getS1Sn+S2Sn−1+S3Sn−2+...+SnS1S21+S22+.....+S2n=n2+9n+142n2+9n+13=1+9n+14n22+9n+13n2∴limn→∞S1Sn+S2Sn−1+S3Sn−2+...+SnS1S21+S22+.....+S2n∴limn→∞n2+9n+142n2+9n+13=1+0+02+0+0=12

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