Why does this hold?

Question

$$\frac{(n+1)(n+1)}{(n+1)(n+2)}=1-\frac{1}{n+2}$$ I have tried WolframAlpha but I don't get any step by step solutions.

Answer 1

Yes. It holds as long as $n$ is a real number different from $-1$ and $-2$. $$\frac{(n+1)(n+1)}{(n+1)(n+2)}=\frac{n+1}{n+2} = \frac{n+1+1-1}{n+2} = \frac{n+2-1}{n+2} = \frac{n+2}{n+2} - \frac{1}{n+2} = 1 - \frac{1}{n+2}.$$

Answer 2

$$\frac{(n+1)(n+1)}{(n+1)(n+2)}=\frac{n+1}{n+2}=\frac{n+2-1}{n+2}=\frac{n+2}{n+2}+\frac{-1}{n+2}=1-\frac{1}{n+2} $$