Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been trying to simplify this $$ 1-\frac{1}{n+2}+\frac{1}{(n+2) (n+3)} $$ to get it to that $$ 1-\frac{(n+3)-1}{(n+2)(n+3)} $$ but I always end up with this $$ 1-\frac{(n+3)+1}{(n+2)(n+3)} $$ Any ideas of where I'm going wrong? Wolfram Alpha gets it to correct form but it doesn't show me the steps (even in pro version)


share|cite|improve this question
up vote 3 down vote accepted

Everywhere there is a minus sign, replace it with plus a negative.

So with your original expression, try instead simplifying $$ 1+\frac{-1}{n+2}+\frac{1}{(n+2) (n+3)} $$ and you should be much less prone to error.

share|cite|improve this answer
Great answer, thanks! – Daniel Wardin May 16 '13 at 16:49

You just have the problem that while $$x-y+z = x-(y-z)$$

you are instead writing:$$x-y+z=x-(y+z)$$

share|cite|improve this answer
This is very useful thing to remember! Unbelievable how I missed such a simple property! Wish I could select 2 answers as best. – Daniel Wardin May 16 '13 at 16:50

Just in case you want to see a full simplification:

\begin{align*} 1-\frac{1}{n+2}+\frac{1}{(n+2)(n+3)} &= \frac{(n+2)(n+3)}{(n+2)(n+3)} - \frac{(n+3)}{(n+2)(n+3)}+\frac{1}{(n+2)(n+3)} \\ &= \frac{(n+2)(n+3)-(n+3)+1}{(n+2)(n+3)}\\ &= \frac{(n^2+5n+6) -n-2 }{(n+2)(n+3)} \\ &= \frac{n^2+4n+4}{(n+2)(n+3)} \\ &= \frac{(n+2)^2}{(n+2)(n+3)} \\ &= \frac{n+2}{n+3} \\ \end{align*} Provided $n\neq -2$.

share|cite|improve this answer
Thanks a lot for the in-depth answer, gives me a look at a new way of working things like that out. Wish I could select multiple answers as best – Daniel Wardin May 16 '13 at 16:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.