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Question

If Sn=1135+1357+1579, then 24S=

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Solution

Tn=1(2n1)(2n+1)(2n+3)=14[1(2n1)(2n+1)1(2n+1)(2n+3)]
Which is the form of 14(VnVn+1)
Sn=14(V1Vn+1)Sn=14[131(2n+1)(2n+3)]
So when denominator become vary large then fraction will be 0
Hence S=14[130]=112

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