Question

best and explained answer would be marked as brilliant.
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Answer 1

We have asked to prove that,

$\sf \dfrac{1}{1+\sqrt{2}} + \dfrac{1}{\sqrt{2}+\sqrt{3}} + \dfrac{1}{\sqrt{3}+\sqrt{4}} + \dfrac{1}{\sqrt{4}+\sqrt{5}} + \dfrac{1}{\sqrt{5}+\sqrt{6}} + \dfrac{1}{\sqrt{6}+\sqrt{7}} + \dfrac{1}{\sqrt{7}+\sqrt{8}} + \dfrac{1}{\sqrt{8}+\sqrt{9}} = 2$

So to solve this question we have to rationalize the denominator of each expression. So let's rationalize!

Expression first!

$:\implies \sf \dfrac{1}{1+\sqrt{2}} \\ \\ :\implies \sf \dfrac{1}{1+\sqrt{2}} \times \dfrac{1-\sqrt{2}}{1-\sqrt{2}} \\ \\ :\implies \sf (a+b)(a-b) = a^2 - b^2 \\ \\ :\implies \sf (1)^{2} - (\sqrt{2})^{2} \\ \\ :\implies \sf 1 - 2 \\ \\ :\implies \sf -1 \\ \\ :\implies \sf \dfrac{-\sqrt{2}}{-1} \\ \\ :\implies -(1-\sqrt{2}) \\ \\ :\implies \sf \sqrt{2} - 1$

Expression 2nd

$:\implies \sf \dfrac{1}{\sqrt{2}+\sqrt{3}} \\ \\ :\implies \sf \dfrac{1}{\sqrt{2}+\sqrt{3}} \times \dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}} \\ \\ :\implies \sf (a+b)(a-b) = a^2 - b^2 \\ \\ :\implies \sf (\sqrt{2})^{2} - (\sqrt{3})^{2} \\ \\ :\implies \sf 2 - 3 \\ \\ :\implies \sf -1 \\ \\ :\implies \sf \dfrac{\sqrt{2}-\sqrt{3}}{-1} \\ \\ :\implies \sf \sqrt{3} - \sqrt{2}$

Expression 3rd

$:\implies \sf \dfrac{1}{\sqrt{3}+\sqrt{4}} \\ \\ :\implies \sf \dfrac{1}{\sqrt{3}+\sqrt{4}} \times \dfrac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}} \\ \\ :\implies \sf (a+b)(a-b) = a^2 - b^2 \\ \\ :\implies \sf (\sqrt{3})^{2} - (\sqrt{4})^{2} \\ \\ :\implies \sf 3 - 4 \\ \\ :\implies \sf -1 \\ \\ :\implies \sf \dfrac{\sqrt{3}-\sqrt{4}}{-1} \\ \\ :\implies \sf \sqrt{4} - \sqrt{3}$

Expression fourth

$:\implies \sf \dfrac{1}{\sqrt{4}+\sqrt{5}} \\ \\ :\implies \sf \dfrac{1}{\sqrt{4}+\sqrt{5}} \times \dfrac{\sqrt{4}-\sqrt{5}}{\sqrt{4}-\sqrt{5}} \\ \\ :\implies \sf (a+b)(a-b) = a^2 - b^2 \\ \\ :\implies \sf (\sqrt{4})^{2} - (\sqrt{5})^{2} \\ \\ :\implies \sf 4 - 5 \\ \\ :\implies \sf 1 \\ \\ :\implies \sf \dfrac{\sqrt{4}-\sqrt{5}}{-1} \\ \\ :\implies \sf \sqrt{5} - \sqrt{4}$

Expression fifth!

$:\implies \sf \dfrac{1}{\sqrt{5}+\sqrt{6}} \\ \\ :\implies \sf \dfrac{1}{\sqrt{5}+\sqrt{6}} \times \dfrac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}} \\ \\ :\implies \sf (a+b)(a-b) = a^2 - b^2 \\ \\ :\implies \sf (\sqrt{5})^{2} - (\sqrt{6})^{2} \\ \\ :\implies \sf 5 - 6 \\ \\ :\implies \sf -1 \\ \\ :\implies \sf \dfrac{\sqrt{5}-\sqrt{6}}{-1} \\ \\ :\implies \sf \sqrt{6} - \sqrt{5}$

Expression sixth!

$:\implies \sf \dfrac{1}{\sqrt{6}+\sqrt{7}} \\ \\ :\implies \sf \dfrac{1}{\sqrt{6}+\sqrt{7}} \times \dfrac{\sqrt{6}-\sqrt{7}}{\sqrt{6}-\sqrt{7}} \\ \\ :\implies \sf (a+b)(a-b) = a^2 - b^2 \\ \\ :\implies \sf (\sqrt{6})^{2} - (\sqrt{7})^{2} \\ \\ :\implies \sf 6 - 7 \\ \\ :\implies \sf -1 \\ \\ :\implies \sf \dfrac{\sqrt{6}-\sqrt{7}}{-1} \\ \\ :\implies \sf \sqrt{7} - \sqrt{6}$

Expression seventh!

$:\implies \sf \dfrac{1}{\sqrt{7}+\sqrt{8}} \\ \\ :\implies \sf \dfrac{1}{\sqrt{7}+\sqrt{8}} \times \dfrac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}} \\ \\ :\implies \sf (a+b)(a-b) = a^2 - b^2 \\ \\ :\implies \sf (\sqrt{7})^{2} - (\sqrt{8})^{2} \\ \\ :\implies \sf 7 - 8 \\ \\ :\implies \sf -1 \\ \\ :\implies \sf \dfrac{\sqrt{7}-\sqrt{8}}{-1} \\ \\ :\implies \sf \sqrt{8} - \sqrt{7}$

Expression eighth!

$:\implies \sf \dfrac{1}{\sqrt{8}+\sqrt{9}} \\ \\ :\implies \sf \dfrac{1}{\sqrt{8}+\sqrt{9}} \times \dfrac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}} \\ \\ :\implies \sf (a+b)(a-b) = a^2 - b^2 \\ \\ :\implies \sf (\sqrt{8})^{2} - (\sqrt{9})^{2} \\ \\ :\implies \sf 8 - 9 \\ \\ :\implies \sf -1 \\ \\ :\implies \sf \dfrac{\sqrt{8}-\sqrt{9}}{-1} \\ \\ :\implies \sf \sqrt{9} - \sqrt{8}$

Now putting the final result of each expression in the whole given expression! Let's see how to do!

$:\implies \sf (\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + (\sqrt{5} - \sqrt{4}) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (\sqrt{8} - \sqrt{7}) + (\sqrt{9} - \sqrt{8})$

Now cancelling the terms.

(Don't forget like terms that have opposite signs can cancel each other like √2 cancel -√2 or etc etc)

After cancelling we are lest with,

$:\implies \sf -1 + \sqrt{9} \\ \\ :\implies \sf -1 + 3 \\ \\ :\implies \sf +2 \\ \\ \sf LHS \: = RHS \\ \\ \sf Henceforth, \: proved$

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