# Prove the following identity by mathematical induction

1. Jul 11, 2008

### Ryuuken

1. The problem statement, all variables and given/known data
Prove the following identity by mathematical induction:
$$\sum_{i=1}^n \frac{1}{(2i - 1)(2i + 1)} = \frac{n}{(2n + 1)}$$

2. Relevant equations

3. The attempt at a solution

Let P(n) = $$\sum_{i=1}^n \frac{1}{(2(1) - 1)(2(1) + 1)} = \frac{1}{(2(1) + 1)}$$

P(1) = $$\sum_{i=1}^n \frac{1}{(1)(3)} = \frac{1}{3}$$ is true

Assuming P(k) is true, then P(k + 1) is also true.

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

Do I just change all the i's and n's to k+1 and expand it until left equation is equal to the right?

2. Jul 11, 2008

### futurebird

Re: Induction

You assume that this is true:
$$\sum_{i=1}^k \frac{1}{(2i - 1)(2i + 1)} = \frac{k}{(2k + 1)}$$

And then find a way to change the above in to this:

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

Notice how I made the replacement everywhere.

3. Jul 11, 2008

### Defennder

Re: Induction

The expression you have there doesn't make sense. The summation is for an index of i, yet k is used. Replacing i with k would still be incorrect. Instead, find a way to write $$\sum^{k+1}_{k=0} \frac{1}{(2k+1)(2k-1)} \ \mbox{in terms of} \ \sum^{k}_{k=0} \frac{1}{(2k+1)(2k-1)} \ \mbox{and the (k+1)th summand}$$ and then use the induction assumption.

PS. Avoid using "i" when you are not working with complex numbers. It gets confusing after a while.