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August 15th, 2014, 07:30 AM   #1
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Roots Denominators in a Series

I encountered this question:

$\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+ \sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1} {\sqrt{1,000,000}+\sqrt{1,000,001}}=\sqrt{a}-\sqrt{b}$
Determine the value of a and b.

How to determine the denominator in that series? On a side note, why are there 2 squareroot codes which can't be displayed properly?

Last edited by greg1313; August 15th, 2014 at 04:05 PM. Reason: Fixed square root symbols
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August 15th, 2014, 08:08 AM   #2
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Your latex code showed \sqr t. There should be no space.

To simplify, rewrite each denominator so that the larger radical is first. Multiply each fraction by its conjugate. Then telescope the sum.
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August 15th, 2014, 08:22 AM   #3
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Quote:
Originally Posted by mrtwhs View Post
Your latex code showed \sqr t. There should be no space.

To simplify, rewrite each denominator so that the larger radical is first. Multiply each fraction by its conjugate. Then telescope the sum.
I am eating right now, so I can't calculate on paper in the current time. However I can imagine that all denominators will be 1, right? Due to $(a-b)^2=(a+b)(a-b)$ formula.
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August 15th, 2014, 03:34 PM   #4
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I have been imagining to solve this question before I went to sleep last night, and the answer is $\sqrt{1,000,001}-\sqrt{2}$, right?
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August 15th, 2014, 03:55 PM   #5
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Yes. well done.

I think it's a better (more elegant) question as
$$\sum_{k=1}^{n^2 - 1}\frac{1}{\sqrt{k} + \sqrt{k+1}}$$
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August 16th, 2014, 01:34 AM   #6
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OK, now on a side note, why weren't the roots displayed properly last night despite I had been checking and I didn't write "\sqr t" but normal "\sqrt"?
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