I think your question is :
3 / 12*22 + 5 / 22*32 + 7 / 32*42 + ......................... upto n terms.
Here, nth term : Tn = (2n + 1) / n2(n + 1)2
= ( 1/n(n + 1) ) ( (2n + 1) / n(n + 1) )
= ( 1/n - 1/(n + 1) ) ( 1/n + 1/(n + 1) )
= ( 1/n2 - 1/(n + 1)2 )
SO : Sum of all the n terms : Sn = 
Tn = 1/12 - 1/(n + 1)2
= 1 - 1/(n + 1)2
Moderation Team
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