# help with equation in alternate form

• Oct 21st 2009, 06:23 PM
absvalue
help with equation in alternate form
I'm told that the following two equations are equal, and asked to write the steps to manipulate the first equation into the second.

$\displaystyle 1 - \frac{1}{k + 1} + \frac{1}{(k+1)(k+2)}$

to

$\displaystyle 1 - \frac{1}{k + 1}\bigg[1 - \frac{1}{(k+2)}\bigg]$

I've tried many different ways of writing the first equation, but I can't seem to get it to work. Could someone point me in the right direction?
• Oct 21st 2009, 06:31 PM
ramiee2010
please don't edit your problem if you type something wrong then type it in another post.
\\edit
just open brackets
• Oct 21st 2009, 07:34 PM
Guess992
Perhaps it would be easier to show you how to get from the bottom to the top instead of the other way around.
$\displaystyle 1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1-\frac{1}{(k+1)} \bigg[1-\frac{1}{(k+2)}\bigg]$

$\displaystyle 1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1 + \bigg[\bigg(-\frac{1}{(k+1)}*1\bigg) - \bigg(-\frac{1}{(k+1)}*\frac{1}{(k+2)}\bigg)\bigg]$

$\displaystyle 1 - \frac{1}{(k+1)} + \frac{1}{(k+1)(k+2)} = 1 + \bigg(-\frac{1}{(k+1)}\bigg) + \bigg(\frac{1}{(k+1)(k+2)}\bigg)$

Does that make it easier? I think the most common error people would make there is not distributing (or removing) the negative along with $\displaystyle \frac{1}{(k+1)}$.
• Oct 21st 2009, 07:38 PM
absvalue
Thanks so much! That makes sense. (Nod)