<s>[INST] Find the sum $\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$ 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?
 [/INST] 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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