$\displaystyle \sum_{r=1}^{n}\frac {2r}{(2r-1)(2r+1)(2r+3)}$

No need show me the method, just give me the answer, thanks.

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- Jan 23rd 2008, 01:01 AMSengNeeSigma notation
$\displaystyle \sum_{r=1}^{n}\frac {2r}{(2r-1)(2r+1)(2r+3)}$

No need show me the method, just give me the answer, thanks. - Jan 23rd 2008, 03:14 AMmr fantastic
From the usual partial fraction decomposition:

$\displaystyle \frac {2r}{(2r-1)(2r+1)(2r+3)} = \frac{1}{8} \left( \frac{1}{2r - 1}\right) + \frac{1}{4} \left( \frac{1}{2r + 1}\right) - \frac{3}{8} \left( \frac{1}{2r + 3}\right)$.

So the for n > 2, the first few terms of the series are:

$\displaystyle \frac{1}{8} \left( 1 + \frac{1}{3} + \frac{1}{5} + ....\right) + \frac{1}{4} \left( \frac{1}{3} + \frac{1}{5} + .....\right) - \frac{3}{8} \left( \frac{1}{5} + \frac{1}{7} + ....\right)$.

Note that $\displaystyle \frac{1}{8} + \frac{1}{4} - \frac{3}{8} = 0$ ......

So what terms will be left if n = 3? n = 4? n = n?

If I was less lazy I'd see this through to the answer, which I would then give you ..... - Jan 23rd 2008, 09:21 AMtopsquark
- Jan 23rd 2008, 06:41 PMSengNee
- Jan 23rd 2008, 06:43 PMSengNee
- Jan 24th 2008, 02:30 AMtopsquark