Maths Olympiad Prep

Library / /396 of 520

Algebra Difficulty 5.7 AIME, harder Prove it

Three. (20 points) Given the sequence {an}\left\{a_{n}\right\}, where
a1=1,an+1=12(an+4an)(n∈N+).  a_{1}=1, a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{4}{a_{n}}\right)\left(n \in \mathbf{N}_{+}\right) \text {. }

Prove: For n⩾2n \geqslant 2, we have an>2+4(13)2n−1a_{n}>2+4\left(\frac{1}{3}\right)^{2^{n-1}}.

Solution

an+1+2=an2+4an+42an=(an+2)22an.an+1−2=an2−4an+42an=(an−2)22an. \begin{array}{l} a_{n+1}+2=\frac{a_{n}^{2}+4 a_{n}+4}{2 a_{n}}=\frac{\left(a_{n}+2\right)^{2}}{2 a_{n}} . \\ a_{n+1}-2=\frac{a_{n}^{2}-4 a_{n}+4}{2 a_{n}}=\frac{\left(a_{n}-2\right)^{2}}{2 a_{n}} . \end{array}

Thus, an+1+2an+1−2=(an+2an−2)2\frac{a_{n+1}+2}{a_{n+1}-2}=\left(\frac{a_{n}+2}{a_{n}-2}\right)^{2}.
Therefore, when n⩾2n \geqslant 2,
an+2an−2=(an−1+2an−1−2)2=(an−2+2an−2−2)22=⋯=(a1+2a1−2)2n−1⇒an+2an−2=32n−1⇒an+2=32n−1(an−2)⇒an=2(1+32n−1)32n−1−1=2+432n−1−1⇒an>2+4(13)2n−1. \begin{array}{l} \frac{a_{n}+2}{a_{n}-2}=\left(\frac{a_{n-1}+2}{a_{n-1}-2}\right)^{2}=\left(\frac{a_{n-2}+2}{a_{n-2}-2}\right)^{2^{2}} \\ =\cdots=\left(\frac{a_{1}+2}{a_{1}-2}\right)^{2^{n-1}} \\ \Rightarrow \frac{a_{n}+2}{a_{n}-2}=3^{2^{n-1}} \\ \Rightarrow a_{n}+2=3^{2^{n-1}}\left(a_{n}-2\right) \\ \Rightarrow a_{n}=\frac{2\left(1+3^{2^{n-1}}\right)}{3^{2^{n-1}}-1}=2+\frac{4}{3^{2^{n-1}}-1} \\ \Rightarrow a_{n}>2+4\left(\frac{1}{3}\right)^{2^{n-1}} . \end{array}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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