Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.6 AIME Find the answer

Let $a_{0}, a_{1}, \ldots$ be a sequence such that $a_{0}=3, a_{1}=2$ , and $a_{n+2}=a_{n+1}+a_{n}$ for all $n \geq 0$ . Find $\sum_{n=0}^{8} \frac{a_{n}}{a_{n+1} a_{n+2}}$

Your answer

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

Solution

We can re-write $\frac{a_{n}}{a_{n+1} a_{n+2}}$ as $\frac{a_{n+2}-a_{n+1}}{a_{n+1} a_{n+2}}=\frac{1}{a_{n+1}}-\frac{1}{a_{n+2}}$ . We can thus re-write the sum as $\left(\frac{1}{a_{1}}-\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{2}}-\frac{1}{a_{3}}\right)+\left(\frac{1}{a_{4}}-\frac{1}{a_{3}}\right)+\ldots+\left(\frac{1}{a_{9}}-\frac{1}{a_{10}}\right)=\frac{1}{a_{1}}-\frac{1}{a_{10}}=\frac{1}{2}-\frac{1}{212}=\frac{105}{212}$

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