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Question

Let the series be 11.3.5+13.5.7+15.7.9+... Deduce the sum of the series upto infinity.

A
19
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B
112
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C
116
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D
18
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Solution

The correct option is B 112
Tn=1(2n1)(2n+1)(2n+3)
Split into partial fractions by the method of suppression by putting 2n=1,1,3 successively.
Tn=18(2n1)28(2n+1)+18(2n+3)
Tn=18[(12n112n+1)(12n+112n+3)]
Now putting n=1,2,3,...n and adding
Sn=18[(112n+1)(1312n+3)]
=18[2n2n+12n3(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

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