1993 UNCO Math Contest II Problems/Problem 8

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Problem

For what integer value of $n$ is the expression $$\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots +\frac{1}{\sqrt{n}+\sqrt{n+1}}$$ equal to $7$ ? (Hint: $(1+\sqrt{2})(1-\sqrt{2})=-1.$)

Solution

Looking at the first fraction, we get that $\frac{1}{\sqrt{1}+\sqrt{2}} = \sqrt{2} - \sqrt{1}$. Moving on, we see an interesting pattern in which the fraction $\frac{1}{\sqrt{n}+\sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}$. This means that we can rewrite the fractions as $(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{n+1} - \sqrt{n})$. We can cancel out most of the terms in that sequence and get $\sqrt{n+1} - \sqrt{1} = \sqrt{n+1} - 1$. However, we need to solve for $n$ now. Setting the previous expression equal to $7$, we get that $\sqrt{n+1} = 8$. Squaring both sides, we get that $n+1 = 64$. Hence, $n$ = $\boxed{63}$.

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