Question

Let $S=\sum _{n=1}^{9999}\frac{1}{(\sqrt{n}+\sqrt{n+1})(\sqrt[4]{4n}+\sqrt[4]{4n+1})}$, then $S$ is equal to

Answer 1

$S=\sum _{n=1}^{9999}\frac{1}{(\sqrt{n}+\sqrt{n+1})(\sqrt[4]{4n}+\sqrt[4]{4n+1})}$

$\Rightarrow S=\sum _{n=1}^{9999}\frac{1\times (\sqrt[4]{4n}-4\sqrt[4]{4n+1})}{(\sqrt{n}+\sqrt{n+1})(\sqrt[4]{4n}+\sqrt[4]{4n+1})\times (\sqrt[4]{4n}-\sqrt[4]{4n+1})}$

$\Rightarrow S=\sum _{n=1}^{9999}\frac{(\sqrt[4]{4n}-\sqrt[4]{4n+1})}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n}-\sqrt{n+1})}$

$\Rightarrow S=\sum _{n=1}^{9999}\sqrt[4]{4n+1}-\sqrt[4]{4n}$

$\Rightarrow S=\sqrt[4]{42}-\sqrt[4]{41}+\sqrt[4]{43}-\sqrt[4]{42}+...+\sqrt[4]{410000}-\sqrt[4]{49999}$
$\Rightarrow S=10-1=9$