Question

Find the value of $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}........+\frac{1}{\sqrt{99}+\sqrt{100}}$

Answer 1

The given series is-

$\frac{1}{(1+\sqrt{2})}+\frac{1}{(\sqrt{2}+\sqrt{3})}+...........\dots \dots \dots \dots \dots \dots ..99$ terms

Rationalizing each term, we get,

$=\frac{(\sqrt{2}-1)}{(2-1)}+\frac{(\sqrt{3}-\sqrt{2})}{(3-2)}+.......\dots \dots \dots \dots \dots \dots ..99$ terms

Therefore the $99th$ term will be $\sqrt{100}-\sqrt{99}$ as the $n^{th}$ term of this series is -

$=\sqrt{(n+1)}-\sqrt{n}$

$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+..........\dots \dots \dots \dots \dots \dots \dots \dots +\sqrt{100}-\sqrt{99}$

Thus, Summation of the series should be-

$=-1+\sqrt{100}$ ( because the other terms cancel out each other)

$=9$

So the answer is $9.$

This is the right answer.