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# $\dfrac{1}{(\sqrt{9}-\sqrt{8})}-\dfrac{1}{(\sqrt{8}-\sqrt{7})}+\dfrac{1}{(\sqrt{7}-\sqrt{6})}-\dfrac{1}{(\sqrt{6}-\sqrt{5})}+\dfrac{1}{(\sqrt{5}-\sqrt{4})}$ is equal to   (A). 0    (B) $\dfrac{1}{3}$.    (C). 1   (D). 5

Last updated date: 24th Jun 2024
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Hint: To solve the given equation we will first rationalize each of the terms given to us separately. After rationalizing each term we will find that the denominator of each term will become 1. Because the denominator becomes 1 we can write numerators separately.
After writing the numerator separately we will proceed accordingly and get the desired answer.

First, we will rationalize each term:
$\Rightarrow \dfrac{1}{(\sqrt{9}-\sqrt{8})}\times \dfrac{(\sqrt{9}+\sqrt{8})}{(\sqrt{9}+\sqrt{8})}$
$\Rightarrow \dfrac{1}{(\sqrt{9}-\sqrt{8})}\times \dfrac{(\sqrt{9}+\sqrt{8})}{(\sqrt{9}+\sqrt{8})}-\dfrac{1}{(\sqrt{8}-\sqrt{7})}\times \dfrac{(\sqrt{8}+\sqrt{7})}{(\sqrt{8}+\sqrt{7})}+$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{1}{(\sqrt{7}-\sqrt{6})}\times \dfrac{(\sqrt{7}+\sqrt{6})}{(\sqrt{7}+\sqrt{6})}-\dfrac{1}{(\sqrt{6}-\sqrt{5})}\times \dfrac{(\sqrt{6}+\sqrt{5})}{(\sqrt{6}+\sqrt{5})}+\dfrac{1}{(\sqrt{5}-\sqrt{4})}\times \dfrac{(\sqrt{5}+\sqrt{4})}{(\sqrt{5}+\sqrt{4})}$

$\Rightarrow\dfrac{(\sqrt{9}+\sqrt{8})}{{{(\sqrt{9})}^{2}}-{{(\sqrt{8})}^{2}}}-\dfrac{(\sqrt{8}+\sqrt{7})}{{{(\sqrt{8})}^{2}}-{{(\sqrt{7})}^{2}}}+\dfrac{(\sqrt{7}+\sqrt{6})}{{{(\sqrt{7})}^{2}}-{{(\sqrt{6})}^{2}}}-\dfrac{(\sqrt{6}+\sqrt{5})}{{{(\sqrt{6})}^{2}}-{{(\sqrt{5})}^{2}}}+\dfrac{(\sqrt{5}+\sqrt{4})}{{{(\sqrt{5})}^{2}}-{{(\sqrt{4})}^{2}}}$

We used the formula $(a+b)(a-b)=({{a}^{2}}-{{b}^{2}})$ and ${{(\sqrt{a})}^{2}}=a$
After rationalizing we will simplify:

$\Rightarrow \dfrac{(\sqrt{9}+\sqrt{8})}{9-8}-\dfrac{(\sqrt{8}+\sqrt{7})}{8-7}+\dfrac{(\sqrt{7}+\sqrt{6})}{7-6}-\dfrac{(\sqrt{6}+\sqrt{5})}{6-5}+\dfrac{(\sqrt{5}+\sqrt{4})}{5-4}$
$\Rightarrow\dfrac{(\sqrt{9}+\sqrt{8})}{1}-\dfrac{(\sqrt{8}+\sqrt{7})}{1}+\dfrac{(\sqrt{7}+\sqrt{6})}{1}-\dfrac{(\sqrt{6}+\sqrt{5})}{1}+\dfrac{(\sqrt{5}+\sqrt{4})}{1}$
$\Rightarrow\sqrt{9}+\sqrt{8}-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5})+(\sqrt{5}+\sqrt{4})$
$\Rightarrow\sqrt{9}+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+\sqrt{4}$
$\Rightarrow \sqrt{9}+\sqrt{4}$
$\Rightarrow 3+2=5\text{ }\!\![\!\!\text{ }\because \sqrt{9}=\sqrt{{{3}^{2}}}\text{=3; }\sqrt{4}=\sqrt{{{2}^{2}}}=2\text{ }\!\!]\!\!\text{ }$

We see that all the terms except for the first and last term get canceled.
Thus, the answer to the given expression will be yielded as options D. 5.
So, the correct answer is “Option D”.

Note: This type of similar question can be expressed in the way$\dfrac{1}{(\sqrt{n}-\sqrt{n-1})}-\dfrac{1}{(\sqrt{n-1}-\sqrt{n-2})}.......+\dfrac{1}{(\sqrt{n-(1-x)}-\sqrt{n-x})}-\dfrac{1}{(\sqrt{n-x}-\sqrt{n-(x+1)})}$

Where only the first and last term remains because there are no terms like$\sqrt{n}$ and $\sqrt{n-(x+1)}$
Thus, the only general method of solving a similar type of sums is by rationalizing the denominator and proceeding with the question accordingly. Whenever there is$(n+(n+1)\text{ }\!\![\!\!\text{ Example: 1+2 }\!\!]\!\!\text{ or }n+(n-1)\,\,\,\,\,\,\,\,\,\,[\text{Example: 4+3}]\text{ , etc)}$ type of equation or anything similar to it in the denominator we should try to bring these values to the denominator either by rationalization or any other method since this will decrease the difficulty of the calculations.
Be careful when opening the brackets after rationalization.