# Proof by Induction

1. Jul 15, 2011

### dystplan

The problem statement, all variables and given/known data
Prove by mathematical induction:
$\sum^n_{i=1} 1/((2i-1)(2i+1)) = n/(2n+1)$

The attempt at a solution
$\sum^k+1_{i=1} = 1/((2k-1)(2k+1)) + 1/((2(k+1)-1)(2(k+1)+1))$

= $k/(2k+1) + 1/(4(k+1)^2 + 2(k+1) - 2(k+1) -1)$

= $k/(2k+1) + 1/(4(k+1)^2 -1)$

= $k/(2k+1) + 1/(4(k^2 + 2k + 1))$

= $k/(2k+1) + 1/(4k^2 8k + 3) - 1)$

Solved (thanks Dick):

= $k/(2k+1) + 1/((2k+3)(2k+1))$

= $k(2k+3)/(2k+3)(2k+1) + 1/((2k+3)(2k+1))$

= $(k(2k+3) + 1)/(2k+3)(2k+1)$

= $(2k^2 + 3k +1)/(2k+3)(2k+1)$

= $((k+1)(2k+1))/(2k+3)(2k+1)$

= $(k+1)/(2k+3)$

= $(k+1)/(2k+2+1)$

= $(k+1)/(2(k+1)+1)$

Last edited: Jul 15, 2011
2. Jul 15, 2011

### Dick

You want to show your result is equal to (k+1)/(2*(k+1)+1). You are going to have a hard time doing that because there's a mistake. 1/(4*(k+1)^2-1) is not equal to 1/(4k^2+2k).

3. Jul 15, 2011

### dystplan

Thanks; I've fixed that problem; durh.

Still have no idea how to get from

1/(4k^2+8k+3)−1) (or 1/((2k+3)(2k+1)) ) to (k+1)/(2*(k+1)+1)

4. Jul 15, 2011

### Dick

You want to go from k/(2k+1)+1/((2k+1)*(2k+3)) to (k+1)/(2*(k+1)+1) or (k+1)/(2k+3). It's just algebra. Put everything over a common denominator. Look for common factors to cancel, etc. You don't need to think hard about it before you try just turning the crank and see what happens.

5. Jul 15, 2011

### dystplan

Thanks for the help Dick.