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09 Nov 2020, 21:47
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C
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Difficulty:
55%
(hard)
Question Stats:
64% (01:44) correct
36%
(01:56)
wrong
based on 387
sessions
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Find the value of expression \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}...+\frac{1}{\sqrt{1599}+\sqrt{1600}}\).
A) 52
B) 41
C) 39
D) 34
E) 28
A) 52
B) 41
C) 39
D) 34
E) 28
09 Nov 2020, 23:37
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gmatcafe
We can use the formula \(a^2-b^2=(a+b)(a-b)\) in each term of the expression, where the denominator is in the from a+b.
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}...+\frac{1}{\sqrt{1599}+\sqrt{1600}}\)
\(\frac{\sqrt{2}-\sqrt{1}}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}...\)
\(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+....\sqrt{1599}-\sqrt{1598}+\sqrt{1600}-\sqrt{1599}\)
\(\sqrt{1600}-\sqrt{1}\)
\(40-1\)
\(39\)
C
A copy of below question
https://gmatclub.com/forum/what-is-the-value-of-the-following-expression-320830.html
General Discussion
04 Aug 2024, 00:08
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↧↧↧ Detailed Video Solution to the Problem Series ↧↧↧
We need to simplify \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}...+\frac{1}{\sqrt{1599}+\sqrt{1600}}\)
In these kind of problems idea is to find a pattern so that we can apply the pattern and reduce the problem to a smaller problem.
Let's start by simplifying
\(\frac{1}{\sqrt{1}+\sqrt{2}}\)
We will be using a concept called as Rationalizing Roots, we will multiply the numerator and the denominator with the conjugate of the denominator
Conjugate of \(\sqrt{1}+\sqrt{2} = \sqrt{1}-\sqrt{2}\)
=> \(\frac{1}{\sqrt{1}+\sqrt{2}}\) * \(\frac{\sqrt{1}-\sqrt{2}}{\sqrt{1}-\sqrt{2}}\)
= \(\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1}+\sqrt{2}) * (\sqrt{1}-\sqrt{2})}\)
= \(\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1^2}-\sqrt{2^2})}\)
= \(\frac{\sqrt{1}-\sqrt{2}}{1-2}\)
= \(\frac{\sqrt{1}-\sqrt{2}}{-1}\)
= \(\sqrt{2}-\sqrt{1}\)
Similarly, we can simplify other terms to reduce the problem to
= \(\sqrt{2}-\sqrt{1} + \sqrt{3}-\sqrt{2} + ... + \sqrt{1599}-\sqrt{1598} + \sqrt{1600}-\sqrt{1599}\)
Alternate terms will cancel out and we will be left with
\(\sqrt{1600}-\sqrt{1}\) = 40 - 1 = 39
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Rationalizing Roots
We need to simplify \(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}...+\frac{1}{\sqrt{1599}+\sqrt{1600}}\)
In these kind of problems idea is to find a pattern so that we can apply the pattern and reduce the problem to a smaller problem.
Let's start by simplifying
\(\frac{1}{\sqrt{1}+\sqrt{2}}\)
We will be using a concept called as Rationalizing Roots, we will multiply the numerator and the denominator with the conjugate of the denominator
Conjugate of \(\sqrt{1}+\sqrt{2} = \sqrt{1}-\sqrt{2}\)
=> \(\frac{1}{\sqrt{1}+\sqrt{2}}\) * \(\frac{\sqrt{1}-\sqrt{2}}{\sqrt{1}-\sqrt{2}}\)
= \(\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1}+\sqrt{2}) * (\sqrt{1}-\sqrt{2})}\)
= \(\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1^2}-\sqrt{2^2})}\)
= \(\frac{\sqrt{1}-\sqrt{2}}{1-2}\)
= \(\frac{\sqrt{1}-\sqrt{2}}{-1}\)
= \(\sqrt{2}-\sqrt{1}\)
Similarly, we can simplify other terms to reduce the problem to
= \(\sqrt{2}-\sqrt{1} + \sqrt{3}-\sqrt{2} + ... + \sqrt{1599}-\sqrt{1598} + \sqrt{1600}-\sqrt{1599}\)
Alternate terms will cancel out and we will be left with
\(\sqrt{1600}-\sqrt{1}\) = 40 - 1 = 39
So, Answer will be C
Hope it helps!
Watch the following video to MASTER Rationalizing Roots
