Find - 11-2-1-2-3-1-3-4-1-4-5-1-5-6-1-6-7-1-7-8-1-8-9

11+√(2)+1√(2)+√(3)+1√(3)+√(4)+1√(4)+√(5)+1√(5)+√(6)+1√(6)+√(7)+1√(7)+√(8)+1√(8)+√(9) =∑^ 8_n=11√(n)+√(n+1) =∑^8_n=1(√(n+1) √(n)) =√(2) √(1)+ ...

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Byju's Answer

Standard VIII

Mathematics

Finding Square Root of a Number

Find : 11+√2...

Question

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

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Solution

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

$= \sum_{n = 1}^{8} \frac{1}{\sqrt{n} + \sqrt{n + 1}}$

$= \sum_{n = 1}^{8} (\sqrt{n + 1} - \sqrt{n})$

$= \sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + \sqrt{5} - \sqrt{4} + . . . . . . + \sqrt{9} - \sqrt{8}$

$= \sqrt{9} - 1$

$= 2$

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Similar questions

$\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} =$ __________.

Q. Prove that :

\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{4} + \sqrt{3}} + \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}} = 2

Q. The value of

\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}}

is.

Q. The value of :

\frac{1}{\sqrt{9} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{1}}

is :

Q. Find the value of:

(1) 3⁶
(2) 6³
(3) 2⁶
(4) 1⁸
(5) 5⁴
(6) 4³
(7) 10⁵
(8) 10⁷
(9) 7⁴
(10) 8²

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Find :11+√2+1√2+√3+1√3+√4+1√4+√5+1√5+√6+1√6+√7+1√7+√8+1√8+√9

11+√2+1√2+√3+1√3+√4+1√4+√5+1√5+√6+1√6+√7+1√7+√8+1√8+√9=∑8n=11√n+√n+1=∑8n=1(√n+1−√n)=√2−√1+√3−√2+√4−√3+√5−√4+......+√9−√8=√9−1=2

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