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I would like to check I have this correct

Find the sum $$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$$ Hint: rationalise the denominators to get a 'telescoping' sum: a sum of terms in which many pairs add up to zero.

I rationalised the denominators to get a series like this: $$\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} +...+\frac{\sqrt{99}-\sqrt{100}}{-1} $$ Which can be written: $$\sqrt{2}-\sqrt{1} + \sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}...+ \sqrt{99}-\sqrt{98} +\sqrt{100}-\sqrt{99}$$ Which is the telescoping sum the question talks about. Most of the terms drop out to leave $$-\sqrt{1} +\sqrt{100} = 9$$

Have I got this correct?

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    $\begingroup$ I think you got it right! $\endgroup$ – Tunococ Aug 10 '13 at 11:12
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    $\begingroup$ Yes what you've done looks right.Plus I also confirmed it using Wolfram Alpha ( link ) (8.999... = 9 ) $\endgroup$ – A Googler Aug 10 '13 at 14:02
  • $\begingroup$ Here you haven't mentioned the limit to your question so i am assuming it to be 'n' let the number of terms be up to n rationalize each term in the expression for example --- (1/(1+sqrt2)) =(sqrt2-sqrt1)/((sqrt2+sqrt1)*(sqrt2-sq… =(sqrt2-sqrt1)/(sqrt2^2-sqrt1^2) =sqrt2*sqrt1.(since sqrt(1) = 1) perform the above step for all terms in your question by doing that we get the expression ------- =sqrt2-sqrt1+sqrt3-sqrt2+sqrt4-sqrt3+..… (all the terms cancel out with its corresponding opposite number,except sqrt1 and sqrt(n+1)) therefore answer is = sqrt(n+1)-1 $\endgroup$ – user90882 Aug 18 '13 at 19:01
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you could also do it by induction

conjecture: $\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{n-1}+\sqrt{n}} = \sqrt{n}-1$ For $n \in [2,3,4 ...]$

For $n=2$:

$$\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2}-1$$

For $n+1$: $$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{n-1}+\sqrt{n}} + \frac{1}{\sqrt{n}+\sqrt{n+1}}$$

$$= \sqrt{n} - 1 + \frac{1}{\sqrt{n}+\sqrt{n+1}}$$

$$= \sqrt{n} - 1 + \sqrt{n+1} - \sqrt{n}$$

$$= \sqrt{n+1}-1$$

$$Q.E.D.$$

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    $\begingroup$ This isn't a different answer; it's a (needless) formalism of the OP's. If you wanted to be totally formal about your proof, you could do it this way, but it takes longer to write out. $\endgroup$ – John Gowers Aug 14 '13 at 18:28

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