# Proof by Induction

Homework Statement
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):

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

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

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

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

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

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

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

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

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## Answers and Replies

Dick
Homework Helper
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).

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).

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)

Dick