The correct option is B 112
Tn=1(2n−1)(2n+1)(2n+3)
Split into partial fractions by the method of suppression by putting 2n=1,−1,−3 successively.
∴Tn=18(2n−1)−28(2n+1)+18(2n+3)
∴Tn=18[(12n−1−12n+1)−(12n+1−12n+3)]
Now putting n=1,2,3,...n and adding
Sn=18[(1−12n+1)−(13−12n+3)]
=18[2n2n+1−2n3(2n+3)]=n4[4n+83(2n+1)(2n+3)]
=n(n+2)3(2n+1)(2n+3)
In order to find S∞ we re-write the above
Sn=n2(1+2n)3.4n2(1+12n)(1+32n)
when n→∞
Sn=112