How does $ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$ reduce to $ 1 - \frac{1}{\sqrt{n+2}}\;$? $$ 1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}$$
Reduces to:
$$ 1 - \frac{1}{\sqrt{n+2}} $$
I have no clue how. What is the exact trick here and how can I practice this? I can't properly google the problem.
 A: $\require{cancel}$I'd us a combination putting over a common denominator and factoring.
Using that $foo = \sqrt{foo}^2$ we can factor $\sqrt{n+1}$ out of $(n+2)\sqrt{n+1} + (n+1)\sqrt{(n+2)}= \color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]$
So
$1 - \frac{1}{\sqrt{n+1}} + \frac{1}{(n+2)\sqrt{(n+1)} + (n+1) \sqrt{(n+2)}}=$
$1 -\frac 1{\color{green}{\sqrt{n+1}}} + \frac 1{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1- \frac {(n+2) + \sqrt{n+1}\sqrt{n+2}}{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}+\frac 1{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {(n+2) + \sqrt{n+1}\sqrt{n+2}-1}{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {(n+1) + \sqrt{n+1}\sqrt{n+2}}{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {\color{green}{\sqrt{n+1}}(\sqrt{n+1} + \sqrt{n+2})}{\color{green}{\sqrt{n+1}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {\cancel{\color{green}{\sqrt{n+1}}}(\sqrt{n+1} + \sqrt{n+2})}{\color{green}{\cancel{\color{green}{\sqrt{n+1}}}}[(n+2) + \sqrt{n+1}\sqrt{n+2}]}=$
$1-\frac {\sqrt{n+1}+\sqrt{n+2}}{(n+2)+\sqrt{n+1}\sqrt{n+2}}=$
$1 -\frac {\sqrt{n+1}+\sqrt{n+2}}{\color{red}{\sqrt{n+2}}(\sqrt{n+2} + \sqrt{n+1})}=$
$1 -\frac {\cancel{\sqrt{n+1}+\sqrt{n+2}}}{\color{red}{\sqrt{n+2}}\cancel{(\sqrt{n+2} + \sqrt{n+1})}}=$
$1 - \frac 1{\color{red}{\sqrt{n+2}}}$
A: HINT
Here is a different approach for the sake of curiosity (as suggested by Paul in the comments).
Precisely, we shall multiply the numerator and the denominator by the conjugate of the denominator:
\begin{align*}
\frac{1}{(n+2)\sqrt{n+1} + (n+1)\sqrt{n+2}} & = \frac{(n+2)\sqrt{n+1} - (n+1)\sqrt{n+2}}{(n + 2)^{2}(n+1) - (n+1)^{2}(n+2)}\\\\
& = \frac{(n+2)\sqrt{n+1} - (n+1)\sqrt{n+2}}{(n+1)(n+2)}\\
& = \frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+2}}
\end{align*}
Can you take it from here?
