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Algebra Difficulty 5.3 AIME, harder Find the answer

3. The sequence a1,a2,…a_{1}, a_{2}, \ldots is defined by the equalities

a1=100,an+1=an+1an,n∈N a_{1}=100, \quad a_{n+1}=a_{n}+\frac{1}{a_{n}}, \quad n \in \mathbb{N}

Find the integer closest to a2013a_{2013}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

# Answer: 118.

Solution.

a20132=(a2012+1a2012)2=a20122+2+1a20122=a20112+2⋅2+1a20112+1a20122=…=a12+2⋅2012+1a12+…+1a20112+1a20122 \begin{aligned} a_{2013}^{2}=\left(a_{2012}+\frac{1}{a_{2012}}\right)^{2}=a_{2012}^{2}+2+\frac{1}{a_{2012}^{2}}=a_{2011}^{2} & +2 \cdot 2+\frac{1}{a_{2011}^{2}}+\frac{1}{a_{2012}^{2}}=\ldots \\ & =a_{1}^{2}+2 \cdot 2012+\frac{1}{a_{1}^{2}}+\ldots+\frac{1}{a_{2011}^{2}}+\frac{1}{a_{2012}^{2}} \end{aligned}

Therefore, on one hand,

a20132>a12+2⋅2012=10000+4024=14024>1182=13924 a_{2013}^{2}>a_{1}^{2}+2 \cdot 2012=10000+4024=14024>118^{2}=13924

on the other hand,

a20132<a12+2⋅2012+20121002<14024+1<118.52 a_{2013}^{2}<a_{1}^{2}+2 \cdot 2012+\frac{2012}{100^{2}}<14024+1<118.5^{2}

Therefore, 118<a2013<118.5118<a_{2013}<118.5, and the nearest integer is 118.

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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.