Sum of Series: 1/(1.3.5)+1/(3.5.7)+1/(5.7.9)+...n Terms

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The discussion centers on summing the series 1/(1.3.5) + 1/(3.5.7) + 1/(5.7.9) with a general term expressed as a_n = 1/((2n-1)(2n+1)(2n+3)). Participants are encouraged to find constants A, B, and C for the partial fraction decomposition of the general term. The series is identified as a telescoping series, which simplifies the computation of the n-th partial sum. A request for detailed solutions and numerical values for A, B, and C indicates some participants are struggling with the problem.
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sum up the series :
1/(1.3.5)+1/(3.5.7)+1/(5.7.9)+.......n terms.
 
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The general term of the series is:

<br /> a_{n} = \frac{1}{(2 n - 1)(2 n + 1)(2 n + 3)}, \ n \ge 1<br />

It can be represented in terms of partial fractions:

<br /> a_{n} = \frac{A}{2 n - 1} + \frac{B}{2 n + 1} + \frac{C}{2 n + 3}<br />

Find A, B and C!

After you do that, notice that the denominator of the second term is just that of the first term evaluated for n + 1 and the third is for n + 2. You can use some trick after that which is pretty common to simplify the n-th partial sum of the series

<br /> S_{n} = \sum_{k = 1}^{n} {a_{k}}, \ n \ge 1<br />
 
Dickfore said:
The general term of the series is:

<br /> a_{n} = \frac{1}{(2 n - 1)(2 n + 1)(2 n + 3)}, \ n \ge 1<br />

It can be represented in terms of partial fractions:

<br /> a_{n} = \frac{A}{2 n - 1} + \frac{B}{2 n + 1} + \frac{C}{2 n + 3}<br />

Find A, B and C!

After you do that, notice that the denominator of the second term is just that of the first term evaluated for n + 1 and the third is for n + 2. You can use some trick after that which is pretty common to simplify the n-th partial sum of the series

<br /> S_{n} = \sum_{k = 1}^{n} {a_{k}}, \ n \ge 1<br />

hey, i tried this way but it does not worked out for me or may be i do not grasp your solution. Please give me a detailed solution
 
lakshya91 said:
hey, i tried this way but it does not worked out for me or may be i do not grasp your solution. Please give me a detailed solution

It's called a "telescoping" series. Why don't you start by showing us what numerical values you obtained for "A", "B" and "C" in the partial fractions expansion. Then we can tell you if you're on the right track or not.
 
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