Question #8e330

May 28, 2017

$2.$

Explanation:

The Expression
$= \frac{1}{\sqrt{2} + \sqrt{1}} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}} + \ldots + \frac{1}{\sqrt{9} + \sqrt{8}} ,$

$= \frac{1}{\sqrt{2} + \sqrt{1}} \times \frac{\sqrt{2} - \sqrt{1}}{\sqrt{2} - \sqrt{1}}$
$+ \frac{1}{\sqrt{3} + \sqrt{2}} \times \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$
$+ \frac{1}{\sqrt{4} + \sqrt{3}} \times \frac{\sqrt{4} - \sqrt{3}}{\sqrt{4} - \sqrt{3}}$
$+ \vdots$
$+ \frac{1}{\sqrt{9} + \sqrt{8}} \times \frac{\sqrt{9} - \sqrt{8}}{\sqrt{9} - \sqrt{8}}$

$= \frac{\sqrt{2} - \sqrt{1}}{2 - 1} + \frac{\sqrt{3} - \sqrt{2}}{3 - 2} + \ldots + \frac{\sqrt{9} - \sqrt{8}}{9 - 8}$

$= \left(\cancel{\sqrt{2}} - 1\right) + \left(\cancel{\sqrt{3}} - \cancel{\sqrt{2}}\right) + \left(\cancel{\sqrt{4}} - \cancel{\sqrt{3}}\right) + \ldots + \left(\sqrt{9} - \cancel{\sqrt{8}}\right)$

$= 3 - 1$

$= 2.$

Enjoy Maths.!

May 28, 2017

The answer is $= 2$

Explanation:

We need

$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

The general term is

${u}_{n} = \frac{1}{\sqrt{n} + \sqrt{n + 1}}$

Simplifying the denominator

${u}_{n} = \frac{1}{\sqrt{n + 1} + \sqrt{n}} \cdot \frac{\sqrt{n + 1} - \sqrt{n}}{\sqrt{n + 1} - \sqrt{n}}$

$= \frac{\sqrt{n + 1} - \sqrt{n}}{n + 1 - n}$

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

Therefore,

$n = 1$, $\implies$, ${u}_{1} = \cancel{\sqrt{2}} - 1$

$n = 2$, $\implies$, ${u}_{2} = \cancel{\sqrt{3}} - \cancel{\sqrt{2}}$

$n = 3$, $\implies$, ${u}_{3} = \cancel{\sqrt{4}} - \cancel{\sqrt{3}}$

$n = 7$, $\implies$, ${u}_{7} = \cancel{\sqrt{8}} - \cancel{\sqrt{7}}$

$n = 8$, $\implies$, ${u}_{8} = \sqrt{9} - \cancel{\sqrt{8}}$

${\sum}_{n = 1}^{8} {u}_{n} = \sqrt{9} - 1 = 3 - 1 = 2$