Puzzles
What is the sum?
What is \(\displaystyle\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}\)?
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Start by rationalising the denominators:
$$\begin{array}{rl}
\frac{1}{\sqrt{n}+\sqrt{n+1}}
&=\frac{1}{\sqrt{n}+\sqrt{n+1}}\times\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}\\
&=\frac{\sqrt{n}-\sqrt{n+1}}{n-(n+1)}\\
&=\sqrt{n+1}-\sqrt{n}\end{array}$$
Therefore:
$$\begin{array}{rl}
\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}
&=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\\&...+(\sqrt{15}-\sqrt{14})+(\sqrt{16}-\sqrt{15})\\
&=-\sqrt{1}+\sqrt{16}\\
&=3
\end{array}
$$