$$\frac{(n+1)(n+1)}{(n+1)(n+2)}=1-\frac{1}{n+2}$$ I have tried WolframAlpha but I don't get any step by step solutions.
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1Perhaps you can immediately see that, at least the terms $(n+1)$ cancel out? Do you see it? – Matti P. Oct 16 '20 at 11:08
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2Assuming $n+1\neq0$, you can cancel $n+1$ from the denominator and numerator. – Rushabh Mehta Oct 16 '20 at 11:08
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Yes. It holds as long as $n$ is a real number different from $-1$ and $-2$. $$\frac{(n+1)(n+1)}{(n+1)(n+2)}=\frac{n+1}{n+2} = \frac{n+1+1-1}{n+2} = \frac{n+2-1}{n+2} = \frac{n+2}{n+2} - \frac{1}{n+2} = 1 - \frac{1}{n+2}.$$
Martingalo
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$$\frac{(n+1)(n+1)}{(n+1)(n+2)}=\frac{n+1}{n+2}=\frac{n+2-1}{n+2}=\frac{n+2}{n+2}+\frac{-1}{n+2}=1-\frac{1}{n+2} $$
Lilla
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