Question

⠀⠀⠀
›› Q87. Simplify: $\it{\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{8}+\sqrt{9}}}$

(a) 1
(b) 2
(c) 234
(d) ∞​

Answer 1

Answer:

Option b

Step-by-step explanation:

Given :-

1/(1+√2) + 1/(√2+√3) +...+1/(√8+√9)

To find:-

Simplify the expression ?

Solution:-

Given that

1/(1+√2) + 1/(√2+√3) +...+1/(√8+√9)

We know that

The Rationalising factor of√a+√b = √a-√b

The Rationalising factor of 1+√2 = 1-√2

The Rationalising factor of√2+√3 = √2-√3

The Rationalising factor of√8+√9= √8-√9

On Rationalising the denominator then

[(1-√2)/(1+√2)(1-√2)]+[(√2-√3)/(√2+√3)(√2-√3)]+...

+ [(√8-√9)/(√8+√9)(√8-√9)]

Since (a+b)(a-b) = a^2-b^2

=>[ (1-√2)/(1-2)]+[(√2-√3)/(2-3)]+...+[(√8-√9)/(8-9)]

=> (1-√2)/(-1)+ (√2-√3)/(-1) + ...+ (√8-√9)/(-1)

=> √2-1+√3-√2++...+√9-√8

=> -1+√9

=> -1+3

=> 3-1

=>2

Answer:-

The value of the given expression is 2

Used formulae:-

• The Rationalising factor of√a+√b = √a-√b

• (a+b)(a-b) = a^2-b^2

Answer 2

Answer:

${\huge{\blue{\texttt{\orange A\red N\green S\pink W\blue E\purple R\red}}}}$

(b) 2

Given

• $\it{\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{8}+\sqrt{9}}}$

To find

• Simplyfying the expression

Solution

★On rationalising

$= > \frac{1}{1 + \sqrt{2} } = \frac{1}{1 + \sqrt{2} } \times \frac{1 - \sqrt{2} }{1 - \sqrt{2} }$

$= > \frac{1 - \sqrt{2} }{ {1}^{2} - { \sqrt{2} }^{2} } = \frac{1 - \sqrt{2} }{ - 1}$

$= > \sqrt{2} - 1$

&

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

$= > \frac{ \sqrt{2} - \sqrt{3} }{2 - 3} = \frac{ \sqrt{2} - \sqrt{3} }{ - 1}$

$= > \sqrt{3 - \sqrt{2} }$

★Similarly

• $= > \frac{1}{ \sqrt{3} + \sqrt{4} } = \sqrt{4} - \sqrt{3}$
• $= > \frac{1}{ \sqrt{4} + \sqrt{5} } = \sqrt{5} - \sqrt{4}$
• $= > \frac{1}{ \sqrt{5} + \sqrt{6} } = \sqrt{6} - \sqrt{5}$
• $= > \frac{1}{ \sqrt{6} + \sqrt{7} } = \sqrt{7} - \sqrt{6}$
• $= > \frac{1}{ \sqrt{7} + \sqrt{8} } = \sqrt{8} - \sqrt{7}$
• $= > \frac{1}{ \sqrt{8} + \sqrt{9} } = \sqrt{9 } - \sqrt{8}$

★So now the equation becomes

$=(√2-1)+(√3-√2)+(√4-√3) + (√5 -√4) +(√6-√5)+(√7-√6) + (√8 - √7) + (√9-√8)</p><p>$

$= - 1 + \sqrt{9}$

$= - 1 + 3 = 2$

★Hence

${\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+....+\dfrac{1}{\sqrt{8}+\sqrt{9}}}$

$= 2$

${\huge{\blue{\texttt{\orange T\red h\green a\pink n\blue k\purple s\red}}}}$