1. ## Sigma notation

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

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

2. Originally Posted by SengNee
$\sum_{r=1}^{n}\frac {2r}{(2r-1)(2r+1)(2r+3)}$

No need show me the method, just give me the answer, thanks.
From the usual partial fraction decomposition:

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

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

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

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

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 .....

3. Originally Posted by SengNee
No need show me the method, just give me the answer, thanks.
As this is a Math *HELP* Forum, I find this comment to be amazing. Why would you just want the answer and not learn how to do it?

-Dan

4. Originally Posted by mr fantastic
From the usual partial fraction decomposition:

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

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

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

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

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 .....
n=n

5. Originally Posted by topsquark
As this is a Math *HELP* Forum, I find this comment to be amazing. Why would you just want the answer and not learn how to do it?

-Dan
Because I had know how to do and I did it step by step, but my answer is different to the book's answer.

6. Originally Posted by SengNee
Because I had know how to do and I did it step by step, but my answer is different to the book's answer.
Then it's probably best to at least post your solution to see who agrees with it.

-Dan