Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

Thanks

share|improve this question

3 Answers 3

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|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|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|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

 
discard

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.