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Question

The sum of the series 13+115+135+163+ upto n terms is Sn. Then 4S equals

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Solution

T1=13=1221T2=115=1421T3=135=1621T4=163=1821
Tn=1(2n)21=1(2n+1)(2n1)Tn=12[1(2n1)1(2n+1)]

So, Sn=Tn
=12[1113]+12[1315]+12[1517]

+12[1(2n1)1(2n+1)]

Sn=12[11(2n+1)]
For very large value of denominator, value of the fraction will be 0
S=12

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