Track / Stage 5 / 365 of 400 #965 of 1964
Problem 965 AIME late Difficulty 5.7 Prove it
Given a 1 = 1 , a n + 1 = a n n + n a a_{1}=1, a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a} a 1 = 1 , a n + 1 = n a n + a n n ⩾ 4 n \geqslant 4 n ⩾ 4 n < a n < n + 1 \sqrt{n}<a_{n}<\sqrt{n+1} n < a n < n + 1
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Official solution 8. First prove that f ( x ) = x n + n x f(x)=\frac{x}{n}+\frac{n}{x} f ( x ) = n x + x n ( 0 , n ) (0, n) ( 0 , n ) n ⩽ a n ⩽ n n − 1 \sqrt{n} \leqslant a_{n} \leqslant \frac{n}{\sqrt{n-1}} n ⩽ a n ⩽ n − 1 n a n < n + 1 a_{n}<\sqrt{n+1} a n < n + 1
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Source: NuminaMath-1.5 ,
licensed Apache-2.0 .
Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.