The value of 11-2-1-2-3-1-3-4-1-4-5-1-5-6-1-6-7-1-7-8-1-8-9 is-

The correct option is D 2 nth term of the equation is, T_n=1√(n)+√(n+1) On multiplying and dividing it with √( n+1) √(n) , We get T_n= ...

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Byju's Answer

Standard X

Mathematics

Sum of Infinite Terms of a GP

The value of ...

Question

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.

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Solution

The correct option is D $2$
$n t h$ term of the equation is, $T_{n} = \frac{1}{\sqrt{n} + \sqrt{n + 1}}$
On multiplying and dividing it with $\sqrt{n + 1} - \sqrt{n}$ , We get
$T_{n} = \frac{\sqrt{n + 1} - \sqrt{n}}{n + 1 - n} = \sqrt{(n + 1)} - \sqrt{n}$
Given series will become,
$- 1 + \sqrt{2} + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} \dots + \sqrt{8} - \sqrt{7} + \sqrt{9} - \sqrt{8}$
$= - 1 + \sqrt{9} = - 1 + 3 = 2$

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Similar questions

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 :

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

Q. 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}}

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. 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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The value of 11+√2+1√2+√3+1√3+√4+1√4+√5+1√5+√6+1√6+√7+1√7+√8+1√8+√9 is.

The correct option is D 2nth term of the equation is, Tn=1√n+√n+1On multiplying and dividing it with √n+1−√n, We getTn=√n+1−√nn+1−n=√(n+1)−√nGiven series will become, −1+√2+√3−√2+√4−√3⋯+√8−√7+√9−√8=−1+√9=−1+3=2

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.