Olympiad Maths Prep

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Problem 1080

National olympiad, first round
Algebra Difficulty 6.1 Prove it

10. (20 points) Given the sequence {an}\left\{a_{n}\right\} satisfies:
a1=1,an+1=1+2an(n=1,2,⋯ ).  a_{1}=1, a_{n+1}=1+\frac{2}{a_{n}}(n=1,2, \cdots) \text {. }

Prove: For any positive integer mm, we have
a1+a2+⋯+a2m>4m−14.  a_{1}+a_{2}+\cdots+a_{2 m}>4 m-\frac{1}{4} \text {. }

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

10. From the conditions, we have
an+1+1=2an(an+1),an+1−2=−1an(an−2). Then an+1+1an+1−2=−2×an+1an−2. \begin{array}{l} a_{n+1}+1=\frac{2}{a_{n}}\left(a_{n}+1\right), \\ a_{n+1}-2=-\frac{1}{a_{n}}\left(a_{n}-2\right) . \\ \text { Then } \frac{a_{n+1}+1}{a_{n+1}-2}=-2 \times \frac{a_{n}+1}{a_{n}-2} . \end{array}

Thus, for any positive integer n n , we have
an+1an−2=(−2)n−1a1+1a1−2=(−2)n.  \frac{a_{n}+1}{a_{n}-2}=(-2)^{n-1} \frac{a_{1}+1}{a_{1}-2}=(-2)^{n} \text {. }

Therefore, an=2+3(−2)n−1(n=1,2,⋯ ) a_{n}=2+\frac{3}{(-2)^{n}-1}(n=1,2, \cdots) .
From this, for any positive integer k k , we have
a2k−1+a2k=(2−322k−1+1)+(2+322k−1)>4−3(122k−1−122k)=4−322k. \begin{array}{l} a_{2 k-1}+a_{2 k}=\left(2-\frac{3}{2^{2 k-1}+1}\right)+\left(2+\frac{3}{2^{2 k}-1}\right) \\ >4-3\left(\frac{1}{2^{2 k-1}}-\frac{1}{2^{2 k}}\right)=4-\frac{3}{2^{2 k}} . \end{array}

Thus, for any positive integer m m , we have
∑i=12mai=a1+a2+∑k=2m(a2k−1+a2k)⩾1+3+∑k=2m(4−322k)=4m−∑k=2m34k>4m−316(1−14)−1=4m−14. \begin{array}{l} \sum_{i=1}^{2 m} a_{i}=a_{1}+a_{2}+\sum_{k=2}^{m}\left(a_{2 k-1}+a_{2 k}\right) \\ \geqslant 1+3+\sum_{k=2}^{m}\left(4-\frac{3}{2^{2 k}}\right)=4 m-\sum_{k=2}^{m} \frac{3}{4^{k}} \\ >4 m-\frac{3}{16}\left(1-\frac{1}{4}\right)^{-1}=4 m-\frac{1}{4} . \end{array}

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