Maths Olympiad Prep

Library / China Mathematical Competition /68 of 136

China Mathematical Competition

Algebra Difficulty 4.9 AIME Prove it China

Sequence {an}\{a_n\} satisfies a1=a2=a3a_1 = a_2 = a_3. Let
bn=an+an+1+an+2(n∈N+). b_n = a_n + a_{n+1} + a_{n+2} \quad (n \in \mathbb{N}_+).
If {bn}\{b_n\} is a geometric sequence with common ratio 33, find the value of a100a_{100}.

Solution

By the condition, we know that bn=b1⋅3n−1=3nb_n = b_1 \cdot 3^{n-1} = 3^n (n∈N+)(n \in \mathbb{N}_+).
Thus,
an+3−an=bn+1−bn=3n+1−3n=2⋅3n(n∈N+). a_{n+3} - a_n = b_{n+1} - b_n = 3^{n+1} - 3^n = 2 \cdot 3^n \quad (n \in \mathbb{N}_+).
Therefore,
a100=a1+∑k=133(a3k+1−a3k−2)=1+∑k=1332⋅33k−2=1+6⋅2733−127−1=1+313(399−1)=3100+1013. \begin{align*} a_{100} &= a_1 + \sum_{k=1}^{33} (a_{3k+1} - a_{3k-2}) \\ &= 1 + \sum_{k=1}^{33} 2 \cdot 3^{3k-2} \\ &= 1 + 6 \cdot \frac{27^{33} - 1}{27 - 1} \\ &= 1 + \frac{3}{13}(3^{99} - 1) = \frac{3^{100} + 10}{13}. \end{align*}

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.