Let S = , then 2S equals

JEE MainMathematicsModerateSequences and Series

Let S =
$\sum_{n = 1}^{9999}\mspace{2mu}\frac{1}{(\sqrt{n} + \sqrt{n + 1})(\sqrt[4]{n} + \sqrt[4]{n + 1})}$ , then 2S equals

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Given,

S = \sum_{n = 1}^{9999}\mspace{2mu}\frac{1}{(\sqrt{n} + \sqrt{n + 1})(\sqrt[4]{n} + \sqrt[4]{n + 1})}

S = \sum_{n = 1}^{9999}\mspace{2mu}\frac{1}{(\sqrt{n} + \sqrt{n + 1})(\sqrt[4]{n} + \sqrt[4]{n + 1})}\left( \frac{\sqrt[4]{n} - \sqrt[4]{n + 1}}{\sqrt[4]{n} - \sqrt[4]{n + 1}} \right)

= \sum_{n = 1}^{9999}\mspace{2mu}\left( (n + 1)^{1/4} - n^{1/4} \right)

= \left( \left( 2^{\frac{1}{4}} - 1 \right) + \left( 3^{\frac{1}{4}} - 2^{\frac{1}{4}} \right) + \left( 4^{\frac{1}{4}} - 3^{\frac{1}{4}} \right) + \ldots + \left( (9999 + 1)^{\frac{1}{4}} - (9999)^{\frac{1}{4}} \right) \right)

= \left( 10^{4} \right)^{\frac{1}{4}} - 1 = 9

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

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