Let S-n-199991-n-n-1-n-n-1- then S is equal to

S=∑^9999_n=11(√(n)+√(n+1))(√(n)+√(n+1)) ⇒ S=∑^9999_n=11×(√(n) 4√(n+1))(√(n)+√(n+1))(√(n)+√(n+1))×(√(n) √(n+1)) ⇒ S=∑^9999_n=1(√(n) √(n+1))(√(n)+√(n+1))(√(n) √ ...

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Byju's Answer

Standard XII

Mathematics

Vn Method

Let S=∑n=1999...

Question

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

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Solution

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

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

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

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

$\Rightarrow S = \sqrt[4]{2} - \sqrt[4]{1} + \sqrt[4]{3} - \sqrt[4]{2} + . . . + \sqrt[4]{10000} - \sqrt[4]{9999}$
$\Rightarrow S = 10 - 1 = 9$

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Let S=9999∑n=11(√n+√n+1)(4√n+4√n+1), then S is equal to

S=9999∑n=11(√n+√n+1)(4√n+4√n+1) ⇒S=9999∑n=11×(4√n−44√n+1)(√n+√n+1)(4√n+4√n+1)×(4√n−4√n+1) ⇒S=9999∑n=1(4√n−4√n+1)(√n+√n+1)(√n−√n+1) ⇒S=9999∑n=14√n+1−4√n ⇒S=4√2−4√1+4√3−4√2+...+4√10000−4√9999 ⇒S=10−1=9

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