# Steps between 2 forms of divisons

• Oct 14th 2012, 05:20 AM
Creephun
Steps between 2 forms of divisons
Hello!

(((l+1)^2)/((2l+1)*(2l+3))) + ((l*(l+1))/(2*(2l+1))) - Wolfram|Alpha

I can't solve, how to get the Alternate form from the Input one on the link. Please somebody help me, step-by step to understand, how do the calculator did it.
Thanks :)
• Oct 14th 2012, 06:09 AM
topsquark
Re: Steps between 2 forms of divisons
Quote:

Originally Posted by Creephun
Hello!

(((l+1)^2)/((2l+1)*(2l+3))) + ((l*(l+1))/(2*(2l+1))) - Wolfram|Alpha

I can't solve, how to get the Alternate form from the Input one on the link. Please somebody help me, step-by step to understand, how do the calculator did it.
Thanks :)

Here's a "road map" for you.

I'm going to relabel your "l" since it looks so much like a "1". I'll call it x.
It looks like
$\frac{(x + 1)^2}{(2x + 1)(2x + 3)} + \frac{x(x + 1)}{2(2x + 1)}$

I'm going to do a lot of factoring here. You can do it by the brute force method, but the factoring approach is a bit easier to see. So let's factor what we can here.

$\left [ \frac{x + 1}{2x + 1} \right ] \cdot \left [ \frac{x + 1}{2x + 3} + \frac{x}{2} \right ]$

Fractions and instructions to simplify. Let's get some common denominators and do another factorization.

$\left [ \frac{x + 1}{2(2x + 1)(2x + 3)} \right ] \cdot \left [ 2x + 2 + 2x^2 + 3x \right ]$

Combining terms and factoring the numerator (on the right side brackets) gives
$\left [ \frac{x + 1}{2(2x + 1)(2x + 3)} \right ] \cdot \left [ (x + 2)(2x + 1) \right ]$

and canceling the common terms we finally get
$\frac{(x + 1)(x + 2)}{2(2x + 3)}$

-Dan
• Oct 14th 2012, 06:48 AM
Creephun
Re: Steps between 2 forms of divisons
Thank you very much!