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# Sum of the series 1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2).....

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Sum of the series 1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2)..... [#permalink]

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06 Nov 2009, 15:30
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Sum of the series 1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2)........1/ (120^1/2 + 121^1/2) is

1.10
2.11
3.12
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06 Nov 2009, 16:42
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virtualanimosity wrote:
Q.sum of the series
1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2)........1/ (120^1/2 + 121^1/2) is

1.10
2.11
3.12

Multiplying denominator and nominator by $$\sqrt{x}-\sqrt{x-1}$$. As in denominator we have $$\sqrt{x}+\sqrt{x-1}$$ we'll get$$x-x+1=1$$.

$$\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+....+\sqrt{121}-\sqrt{120}=-\sqrt{1}+\sqrt{121}=-1+11=10$$
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31 Aug 2012, 07:01
Hi Bunuel,

I want to know whether such type of questions are just brain teasers or such questions have actually appeared in GMAT.
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31 Aug 2012, 08:52
fameatop wrote:
Hi Bunuel,

I want to know whether such type of questions are just brain teasers or such questions have actually appeared in GMAT.

This is not a GMAT question, since there are only three answer choices instead of five. Though the concept of rationalisation of a fraction used to solve this problem is definitely tested on the GMAT (rationalisation is performed to eliminate irrational expression in the denominator). For example:

if-x-0-then-106291.html
if-n-is-positive-which-of-the-following-is-equal-to-31236.html
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19 Sep 2012, 05:23
Bunuel wrote:
virtualanimosity wrote:
Q.sum of the series
1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2)........1/ (120^1/2 + 121^1/2) is

1.10
2.11
3.12

Multiplying denominator and nominator by $$\sqrt{x}-\sqrt{x-1}$$. As in denominator we have $$\sqrt{x}+\sqrt{x-1}$$ we'll get$$x-x+1=1$$.

$$\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+....+\sqrt{121}-\sqrt{120}=-\sqrt{1}+\sqrt{121}=-1+11=10$$

How did you arrive at this: $$\sqrt{x}+\sqrt{x-1}$$ ? Since this, for me, looks like: (from stem): 1/ (2^1/2 + 3^1/2) ---> $$\sqrt{x}+\sqrt{x+1}$$.

But I think that since a+b = b+a, then I'm wondering about the first term. When figuring out series, should we take starting point in the 1st term and try to generalize from this? And how would the series have looked like if the questions was --> 1/(2^1/2- 1^1/2) + 1/ (2^1/2 - 3^1/2)........1/ (120^1/2 - 121^1/2) ?

Thanks.
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Re: Sum of the series 1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2)..... [#permalink]

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16 Dec 2013, 09:06
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Sum of the series 1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2)..... [#permalink]

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04 Jul 2014, 09:02
Bunuel wrote:
virtualanimosity wrote:
Q.sum of the series
1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2)........1/ (120^1/2 + 121^1/2) is

1.10
2.11
3.12

Multiplying denominator and nominator by $$\sqrt{x}-\sqrt{x-1}$$. As in denominator we have $$\sqrt{x}+\sqrt{x-1}$$ we'll get$$x-x+1=1$$.

$$\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+....+\sqrt{121}-\sqrt{120}=-\sqrt{1}+\sqrt{121}=-1+11=10$$

Question : How do we know to multiply by sqrt x - sqrt x - 1 ? also, how do we get x - x + 1, can you show the multiplying out, I mess up when i try it. Lastly, how do we get sqrt 2 - sqrt 1 and so on?
Sum of the series 1/(2^1/2+ 1^1/2) + 1/ (2^1/2 + 3^1/2).....   [#permalink] 04 Jul 2014, 09:02
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