Question

Prove that :
$\frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{4} } + ... + \frac{1}{ \sqrt{8} + 3 } = 2$
​

Answer 1

Step-by-step explanation:

Given:-

$\mathrm{Prove \: that:} \frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{4} } + ... + \frac{1}{ \sqrt{8} + 3 } = 2$

What to do:-

To prove that LHS = RHS.

Solution:-

Let's solve the problem

We have,

$\mathrm{\frac{1}{1 + \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} + \sqrt{4} } + ... upto+ \frac{1}{ \sqrt{8} + 3 } = 2}$

[Hint: First we have to rationalise the denominator all the given terms, one by one, and then after rationalised the denominator, We arrange all the value according to the given question, and simplify that all and after simplifyiny then last we get the answer.

Let's go!

Rationalising each term, we get

$\mathrm{First \: term:} \: \frac{1}{1 + \sqrt{2} }$

$⟹ \frac{1}{ \sqrt{2} + 1 } \times \frac{\sqrt{2} - 1 }{ \sqrt{2} - 1 }$

$⟹ \frac{1(\sqrt{2} - 1) }{( \sqrt{2} + 1)( \sqrt{2} - 1 )}$

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

$⟹ \frac{ \sqrt{2} - 1}{( \sqrt{2} {)}^{2} - (1 {)}^{2} }$

$⟹ \frac{ \sqrt{2} - 1 }{2 - 1}$

$⟹ \frac{ \sqrt{2} - 1 }{1}$

$⟹ \red{ \sqrt{2} - 1}$

$\mathrm{Second \: term:} \: \frac{1}{ \sqrt{2} + \sqrt{3} }$

$⟹ \frac{1}{\sqrt{3} + \sqrt{2} } \times \frac{\sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

$⟹ \frac{1( \sqrt{3} - \sqrt{2} ) }{( \sqrt{3} + \sqrt{2} )( \sqrt{3} - \sqrt{2} )}$

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

$⟹ \frac{ \sqrt{3} - \sqrt{2} }{( \sqrt{3} {)}^{2} - ( \sqrt{2} {)}^{2} }$

$⟹ \frac{ \sqrt{3} - \sqrt{2} }{3 - 2}$

$⟹ \frac{ \sqrt{3} - \sqrt{2} }{1}$

$⟹ \red{ \sqrt{3} - \sqrt{2} }$

$\mathrm{Third \: term:} \: \frac{1}{ \sqrt{3} + \sqrt{4} }$

$⟹ \frac{1}{ 2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$⟹ \frac{1(2 - \sqrt{3} )}{(2 + \sqrt{3} )(2 - \sqrt{3}) }$

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

$⟹ \frac{2 - \sqrt{3} }{(2 {)}^{2} - ( \sqrt{3} {)}^{2} }$

$⟹ \frac{2 - \sqrt{3} }{4 - 3}$

$⟹ \frac{2 - \sqrt{3} }{1}$

$⟹ \red{2 - \sqrt{3} }$

$\mathrm{Eight \: term:} \: \frac{1}{ \sqrt{8} + 3}$

$⟹ \frac{1}{ \sqrt{8} + 3 } \times \frac{ \sqrt{8} - 3 }{ \sqrt{8} - 3 }$

$⟹ \frac{1( \sqrt{8} - 3) }{( \sqrt{8} + 3)( \sqrt{8} - 3)}$

$⟹ \frac{ \sqrt{8} - 3 }{( \sqrt{8} {)}^{2} - (3 {)}^{2} }$

$⟹ \frac{ \sqrt{8} - 3 }{8 - 9}$

$⟹ \frac{ \sqrt{8} - 3 }{1}$

$⟹ \red{ \sqrt{8} - 3 }$

Now, arranging all the rationalised denominator according to the given question and simplifyiny that.

$\mathrm{∴ Given \: expression} =( \sqrt{2} - 1) + ( \sqrt{3} - \sqrt{2} ) + (2 - \sqrt{3} ) + ... + \sqrt{8} - 3$

$= 2 = 2$

LHS = RHS

Hence, proved:

:)