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10 Apr 2020, 08:54
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (01:35) correct
42%
(01:52)
wrong
based on 696
sessions
History
Date
Time
Result
Not Attempted Yet
What is the value of the following expression:
\(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{120} + \sqrt{121}} \)
A. 1
B. 9
C. 10
D. 11
E. 12
M37-54
\(\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{120} + \sqrt{121}} \)
A. 1
B. 9
C. 10
D. 11
E. 12
M37-54
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10 Apr 2020, 08:57
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Bunuel
Rationalize by multiplying the denominator and the numerator of each fraction by \(\sqrt{x}-\sqrt{x-1}\) (this algebraic manipulation is called rationalization and is performed to eliminate irrational expression in the denominator). As in the denominator we have \(\sqrt{x}+\sqrt{x-1}\) we'll get \((\sqrt{x}+\sqrt{x-1})(\sqrt{x}-\sqrt{x-1})=x-(x-1)=1\).
We'll be left with the following:
\((\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+....+(\sqrt{121}-\sqrt{120})=-\sqrt{1}+\sqrt{121}=-1+11=10\)
Answer: C
11 Apr 2020, 06:39
Kudos
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When I have a look at this type of a sequence, I tell myself that I have to look for a pattern. In this question, it’s very easy to identify a pattern since there are surds i.e. irrational numbers in the denominator.
When there is a surd (an irrational number) in the denominator, we try to rationalize it by multiplying and dividing each fraction by the conjugate of the surd.
That is, if we have a term √a+ √b in the denominator, we multiply and divide the fraction by √a- √b. When we do this, the denominator evolves into an expression in the form of (a-b) (a+b) which can be re-written as \(a^2 – b^2\) due to which the roots get cancelled out and we have a rational number.
Therefore, \(\frac{1}{ √1+ √2}\) can be rationalized to \(\frac{√2- √1 }{ 2 – 1}\) = √2- √1.
In each term, the numbers given in the denominator differ by 1. We can conclude that all the denominators will give us 1 upon rationlisation.
The sequence can be simplified to look like the one below:
√2- √1 + √3- √2 + √4- √3 + ……. + √121- √120. This lets us cancel off the positive and the negative terms leaving us with √121- √1 = 11 – 1 = 10.
The correct answer option is C. Bunuel, kindly change the OA for this question.
Hope that helps!
When there is a surd (an irrational number) in the denominator, we try to rationalize it by multiplying and dividing each fraction by the conjugate of the surd.
That is, if we have a term √a+ √b in the denominator, we multiply and divide the fraction by √a- √b. When we do this, the denominator evolves into an expression in the form of (a-b) (a+b) which can be re-written as \(a^2 – b^2\) due to which the roots get cancelled out and we have a rational number.
Therefore, \(\frac{1}{ √1+ √2}\) can be rationalized to \(\frac{√2- √1 }{ 2 – 1}\) = √2- √1.
In each term, the numbers given in the denominator differ by 1. We can conclude that all the denominators will give us 1 upon rationlisation.
The sequence can be simplified to look like the one below:
√2- √1 + √3- √2 + √4- √3 + ……. + √121- √120. This lets us cancel off the positive and the negative terms leaving us with √121- √1 = 11 – 1 = 10.
The correct answer option is C. Bunuel, kindly change the OA for this question.
Hope that helps!
