Simplify $\dfrac{1}{\sqrt 1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\dfrac{1}{\sqrt{8}+\sqrt{9}}$
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The first term has $2$. – Nosrati Sep 26 '17 at 14:49
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I editted it already sir – user484792 Sep 26 '17 at 14:53
2 Answers
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Hint:)
Use $$\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=-\sqrt{n}+\sqrt{n+1}$$ and make a telescopic expression.
Nosrati
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Hint:
$$\frac1{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n}-\sqrt{n+1})}=\sqrt{n+1}-\sqrt{n}$$
Then $$\frac{1}{\sqrt 1+\sqrt{2}}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+\sqrt4}+\cdots +\frac{1}{\sqrt8+\sqrt9}=$$ $$=\sqrt2-\sqrt1+\sqrt3-\sqrt2+\sqrt4-\sqrt3+...+\sqrt9-\sqrt8=\sqrt9-\sqrt1=3-1=2$$
Roman83
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The denominator is incorrect. It should be $\sqrt{n+1}-\sqrt n$ in the second radical. – John Lou Sep 26 '17 at 14:59