# Mathematical Induction question

• July 26th 2009, 08:21 PM
Evil Colt
Mathematical Induction question
Hey there guys i was wondering if someone would be able to help me out. I have to get this done by tomorrow at 4:00PM so the faster the better. Here is the problem,
by mathermatical induction prove

n Sigma i=1 1/(2i-1)(2i+1) = n/2n+1

As you know the n is on top of the sigma and the i=1 at the bottom of the sigma

• July 26th 2009, 08:59 PM
mr fantastic
Quote:

Originally Posted by Evil Colt
Hey there guys i was wondering if someone would be able to help me out. I have to get this done by tomorrow at 4:00PM so the faster the better. Here is the problem,
by mathermatical induction prove

n Sigma i=1 1/(2i-1)(2i+1) = n/2n+1

As you know the n is on top of the sigma and the i=1 at the bottom of the sigma

I assume you can do the first two steps. Then the trouble is in step 3, showing that the result is true for n + 1 assuming it's true for n.

So it boils down to showing that $\frac{n}{2n+1} + \frac{1}{(2[n+1] - 1)(2[n+2] + 1)} = \frac{n+1}{2(n+1) + 1}$.
• July 26th 2009, 09:05 PM
Gamma
I think you can verify the base case for n=1.

So you assume the equality holds up to k. You must show it holds for k+1 for the induction to be complete.

$\sum_{i=1}^{k+1}\frac{1}{(2i-1)(2i+1)}=\sum_{i=1}^{k}\frac{1}{(2i-1)(2i+1)}+\frac{1}{(2(k+1)-1)(2(k+1)+1)}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}=\frac{k(2k+ 3)+1}{(2k+1)(2k+3)}=\frac{(2k+1)(k+1)}{(2k+1)(2k+3 )}=\frac{(k+1)}{(2k+3)}
$
$=\frac{(k+1)}{2(k+1)+1}$

this completes the induction.