Prove the following identity by mathematical induction

[Ryuuken]

Homework Statement

Prove the following identity by mathematical induction:
\sum_{i=1}^n \frac{1}{(2i - 1)(2i + 1)} = \frac{n}{(2n + 1)}

Homework Equations

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?

 

[futurebird]

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 into 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.

 

[Defennder]

\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?

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.