Question

Prove that

$\frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{4} } + \frac{1}{ \sqrt{4} + \sqrt{5} } + \frac{1}{ \sqrt{5} + \sqrt{6} } + \frac{1}{ \sqrt{6} + \sqrt{7} } + \frac{1}{ \sqrt{7} + \sqrt{8} } + \frac{1}{ \sqrt{8} + \sqrt{9} }$
prove that this is equal to 2..

​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{4} } + \frac{1}{ \sqrt{4} + \sqrt{5} } + \frac{1}{ \sqrt{5} + \sqrt{6} } + \\ \frac{1}{ \sqrt{6} + \sqrt{7} } + \frac{1}{ \sqrt{7} + \sqrt{8} } + \frac{1}{ \sqrt{8} + \sqrt{9} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Consider,

$\rm :\longmapsto\:\dfrac{1}{1 + \sqrt{2} }$

can be rewritten as

$\rm \: = \: \:\dfrac{1}{ \sqrt{2} + 1}$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{2} + 1} \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1}$

We know,

$\red{\boxed{ \bf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

we get now,

$\rm \: = \:\dfrac{ \sqrt{2} - 1}{ {( \sqrt{2} )}^{2} - {1}^{2} }$

$\rm \: = \: \: \sqrt{2} - 1$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{2} + \sqrt{3} }$

$\rm \: = \: \:\dfrac{1}{ \sqrt{3} + \sqrt{2} }$

On Rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$\rm \: = \: \:\dfrac{ \sqrt{3} - \sqrt{2} }{3 - 2}$

$\rm \: = \: \: \sqrt{3} - \sqrt{2}$

Now, Consider,

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{3} + \sqrt{4} }$

$\rm \: = \: \:\dfrac{1}{ \sqrt{4} + \sqrt{3} }$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{4} + \sqrt{3} } \times \dfrac{ \sqrt{4} - \sqrt{3} }{ \sqrt{4} - \sqrt{3} }$

$\rm \: = \: \:\dfrac{ \sqrt{4} - \sqrt{3} }{4 - 3}$

$\rm \: = \: \: \sqrt{4} - \sqrt{3}$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{4} + \sqrt{5} }$

$\rm \: = \: \:\dfrac{1}{ \sqrt{5} + \sqrt{4} }$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{5} + \sqrt{4} } \times \dfrac{ \sqrt{5} - \sqrt{4} }{ \sqrt{5} - \sqrt{4} }$

$\rm \: = \: \:\dfrac{ \sqrt{5} - \sqrt{4} }{5 - 4}$

$\rm \: = \: \: \sqrt{5} - \sqrt{4}$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{5} + \sqrt{6} }$

$\rm \: = \: \:\dfrac{1}{ \sqrt{6} + \sqrt{5} }$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{6} + \sqrt{5} } \times \dfrac{ \sqrt{6} - \sqrt{5} }{ \sqrt{6} - \sqrt{5} }$

$\rm \: = \: \:\dfrac{ \sqrt{6} - \sqrt{5} }{6 - 5}$

$\rm \: = \: \: \sqrt{6} - \sqrt{5}$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{6} + \sqrt{7} }$

$\rm \: = \: \:\dfrac{1}{ \sqrt{7} + \sqrt{6} }$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{7} + \sqrt{6} } \times \dfrac{ \sqrt{7} - \sqrt{6} }{ \sqrt{7} - \sqrt{6} }$

$\rm \: = \: \:\dfrac{ \sqrt{7} - \sqrt{6} }{7 - 6}$

$\rm \: = \: \: \sqrt{7} - \sqrt{6}$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{7} + \sqrt{8} }$

$\rm \: = \: \:\dfrac{1}{ \sqrt{8} + \sqrt{7} }$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{8} + \sqrt{7} } \times \dfrac{ \sqrt{8} - \sqrt{7} }{ \sqrt{8} - \sqrt{7} }$

$\rm \: = \: \:\dfrac{ \sqrt{8} - \sqrt{7} }{8 - 7}$

$\rm \: = \: \: \sqrt{8} - \sqrt{7}$

Now, Consider

$\rm :\longmapsto\:\dfrac{1}{ \sqrt{8} + \sqrt{9} }$

$\rm \: = \: \:\dfrac{1}{ \sqrt{9} + \sqrt{8} }$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{9} + \sqrt{8} } \times \dfrac{ \sqrt{9} - \sqrt{8} }{ \sqrt{9} - \sqrt{8} }$

$\rm \: = \: \:\dfrac{ \sqrt{9} - \sqrt{8} }{9 - 8}$

$\rm \: = \: \: \sqrt{9} - \sqrt{8}$

Now, Let consider the given expression,

$\rm :\longmapsto\:\frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{4} } + \frac{1}{ \sqrt{4} + \sqrt{5} } + \frac{1}{ \sqrt{5} + \sqrt{6} } + \\ \frac{1}{ \sqrt{6} + \sqrt{7} } + \frac{1}{ \sqrt{7} + \sqrt{8} } + \frac{1}{ \sqrt{8} + \sqrt{9} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

On substituting the values, evaluated above

$\rm \: = \cancel {\sqrt{2}} - 1 + \cancel {\sqrt{3} }- \cancel {\sqrt{2} }+ \cancel {\sqrt{4}} - \cancel {\sqrt{3}} + \cancel {\sqrt{5}} - \cancel {\sqrt{4}} + \\ \cancel {\sqrt{6}} - \cancel {\sqrt{5}} + \cancel {\sqrt{7}} - \cancel {\sqrt{6}} + \cancel {\sqrt{8}} - \cancel {\sqrt{7}} + \sqrt{9} - \cancel {\sqrt{8} }$

$\rm \: = \: \: \sqrt{9} - 1$

$\rm \: = \: \:3 - 1$

$\rm \: = \: \:2$