Wednesday, 8 August 2018

algebra precalculus - Find the sum to n terms of the series $frac{1}{1.2.3}+frac{3}{2.3.4}+frac{5}{3.4.5}+frac{7}{4.5.6}+cdots$..

Question :




Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$



What I have done :



nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.



Therefore the nth term of given series is :



$\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)




By using partial fraction :



and solving for A,B and C we get A = 1/2, B = -1, C =1/2



Putting the values of A,B and C in (1) we get :



$\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$



But by putting $r =1,2,3, \cdots$ I am not getting the answer. Please guide how to solve this problem . Thanks.

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