Question

show 1/√4+√5+1/√5+√6+1/√6+√7+√8+1/√8+√9=1​

Answer 1

Answer :-

◾We have to show that

$\frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 1$

◾Now as we know we that

→ a + b = b + a

◾We can say that

$\frac{1}{\sqrt{5} + \sqrt{4}} + \frac{1}{\sqrt{6} + \sqrt{5}} + \frac{1}{\sqrt{7} + \sqrt{6}} + \frac{1}{\sqrt{8} + \sqrt{7}} + \frac{1}{\sqrt{9} + \sqrt{8}} = 1$

$\implies \sum\limits_{k = 4}^{k= 8} \frac{1}{\sqrt{k+1} + \sqrt{k}}$

◾On rationalizing

$\implies \sum\limits_{k = 4}^{k= 8} \frac{1}{\sqrt{k+1} + \sqrt{k}} \times \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}}$

$\implies \sum\limits_{k = 4}^{k= 8} \frac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k+1})^2 - (\sqrt{k})^2}$

$\implies \sum\limits_{k = 4}^{k= 8} \frac{\sqrt{k+1} - \sqrt{k}}{k+1 - k}$

$\implies \sum\limits_{k = 4}^{k= 8} \frac{\sqrt{k+1} - \sqrt{k}}{1 }$

◾Now the number

$= \tiny {( \sqrt{5} - \sqrt{4} ) + ( \sqrt{6} - \sqrt{5} )+ ( \sqrt{7} - \sqrt{6} )+ ( \sqrt{8} - \sqrt{7} )+ ( \sqrt{9} - \sqrt{8} )}$

$= -\sqrt{4} + \sqrt{9}$

$= \sqrt{9} - \sqrt{4}$

$= 3 - 2$

$= 1$

= LHS

▪️And as

⏩LHS = RHS = 1

◾Hence Shown