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Algebra Difficulty 5.2 AIME, harder Prove it China

Let a0,a1,a2,…,an,…a_0, a_1, a_2, \dots, a_n, \dots be a sequence of numbers satisfying (3−an+1)⋅(6+an)=18(3 - a_{n+1}) \cdot (6 + a_n) = 18, and a0=3a_0 = 3. Then ∑i=0n1ai\sum_{i=0}^{n} \frac{1}{a_i} equals ________.

Solution

Set bn=1anb_n = \frac{1}{a_n}, n=0,1,2,…n = 0, 1, 2, \dots, then (3−1bn+1)(6+1bn)=18(3 - \frac{1}{b_{n+1}})(6 + \frac{1}{b_n}) = 18, namely,
3bn+1−6bn−1=0. 3b_{n+1} - 6b_n - 1 = 0.
Hence bn+1=2bn+13b_{n+1} = 2b_n + \frac{1}{3}, or bn+1+13=2(bn+13)b_{n+1} + \frac{1}{3} = 2(b_n + \frac{1}{3}). So {bn+13}\{b_n + \frac{1}{3}\} is a geometric progression with common ratio 22. Thus
bn+13=2n(b0+13)=2n(1a0+13)=13×2n+1,bn=13(2n+1−1). b_n + \frac{1}{3} = 2^n \left(b_0 + \frac{1}{3}\right) = 2^n \left(\frac{1}{a_0} + \frac{1}{3}\right) = \frac{1}{3} \times 2^{n+1}, \\ b_n = \frac{1}{3}(2^{n+1} - 1).

∑i=0n1ai=∑i=0nbi=∑i=0n13(2i+1−1)=13[2(2n+1−1)2−1−(n+1)] \sum_{i=0}^{n} \frac{1}{a_i} = \sum_{i=0}^{n} b_i = \sum_{i=0}^{n} \frac{1}{3}(2^{i+1} - 1) = \frac{1}{3} \left[ \frac{2(2^{n+1} - 1)}{2-1} - (n+1) \right]
=13(2n+2−n−3). = \frac{1}{3}(2^{n+2} - n - 3).

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.