Wednesday, 31 July 2013

calculus - Sum to n terms the series frac13cdot9cdot11+frac15cdot11cdot13+frac17cdot13cdot15+cdots.




Q:Sum to n terms the series :
13911+151113+171315+





This was asked under the heading using method of difference and ans given was
Sn=1140148(12n+3+12n+5+12n+732n+9)




My Approach:First i get Un=1(2n+1)(2n+7)(2n+9)
In order to make Un is the reciprocal of the product of factors in A.P i rewrite it
Un=(2n+3)(2n+5)(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)=(2n+7)(2n+9)48(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)=1(2n+1)(2n+3)(2n+5)48(2n+1)(2n+3)(2n+5)(2n+7)(2n+9)
Then i tried to make Un=VnVn1 in order to get Sn=VnV0.But i really don't know how can i figure out this.Any hints or solution will be appreciated.
Thanks in advance.




Answer



Let
Sn=nk=11(2k+1)(2k+7)(2k+9).
By the partial fraction decomposition:
1(2k+1)(2k+7)(2k+9)=1/482k+11/122k+7+1/162k+9
Then, after letting On=nk=112k+1, we have that
Sn=On48On+3O312+On+4O416=4O33O448+On4On+3+3On+448=1140148(On+3On3(On+4On+3))=1140148(12n+3+12n+5+12n+732n+9).


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