Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Find the answer

Example 7 Calculate 121+2+132+23\frac{1}{2 \sqrt{1}+\sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}
+143+34+⋯+110099+99100. +\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\cdots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}} .

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

 Solution: ∵1(n+1)n+nn+1=1n⋅n+1(n+n+1)=n+1−nn⋅n+1=1n−1n+1,∴ the original expression =(11−12)+(12−13)+(13−14)+⋯+(199−1100)=1−110=910.\begin{array}{l}\text { Solution: } \because \frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}} \\ =\frac{1}{\sqrt{n} \cdot \sqrt{n+1}(\sqrt{n}+\sqrt{n+1})} \\ =\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n} \cdot \sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}, \\ \therefore \text { the original expression }=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right) \\ +\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+\cdots+\left(\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right) \\ =1-\frac{1}{10}=\frac{9}{10}. \\\end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.