# 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
Science Advisor
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
Science Advisor
Homework Helper
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.

Thanks for the help Dick.