Q:Sum to n terms the series :
13⋅9⋅11+15⋅11⋅13+17⋅13⋅15+⋯
This was asked under the heading using method of difference and ans given was
Sn=1140−148(12n+3+12n+5+12n+7−32n+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=Vn−Vn−1 in order to get Sn=Vn−V0.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=n∑k=11(2k+1)(2k+7)(2k+9).
By the partial fraction decomposition:
1(2k+1)(2k+7)(2k+9)=1/482k+1−1/122k+7+1/162k+9
Then, after letting On=∑nk=112k+1, we have that
Sn=On48−On+3−O312+On+4−O416=4O3−3O448+On−4On+3+3On+448=1140−148(On+3−On−3(On+4−On+3))=1140−148(12n+3+12n+5+12n+7−32n+9).
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