Find the sum $\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{99}+\sqrt{100}}$
Solution 1:
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 $n+1$: $$\frac{1}{\sqrt{1}+\sqrt{2}} + \frac{1}{\sqrt{2}+\sqrt{3}} + ...+ \frac{1}{\sqrt{n-1}+\sqrt{n}} + \frac{1}{\sqrt{n}+\sqrt{n+1}}$$
$$= \sqrt{n} - 1 + \frac{1}{\sqrt{n}+\sqrt{n+1}}$$
$$= \sqrt{n} - 1 + \sqrt{n+1} - \sqrt{n}$$
$$= \sqrt{n+1}-1$$
$$Q.E.D.$$