2023 SSMO Relay Round 1 Problems/Problem 2

Problem

Let $T=TNYWR$ . Let $a_0 = 3, a_1 = 1, a_2 = T$ , and let $a_n = a_{n-1} - \frac{a_{n-3}}{8}$ for $n\ge 3.$ Find $\sum_{i=0}^\infty a_i.$

Solution

We have $T = 2022$ . Let $\begin{align*} S = \sum_{i=0}^\infty a_i &= a_0+a_1+a_2+\sum_{i=3}^\infty a_i\\ &= 3+1+2022+\sum_{i=3}^\infty \left(a_{i-1}-\frac{a_{i-3}}{8}\right)\\ &= 2026+\sum_{i=3}^{\infty}a_{i-1}-\sum_{i=3}^\infty \frac{a_{i-3}}{8}\\ &= 2026+\sum_{i=2}^\infty a_i-\frac{\sum_{i=0}a_i}{8}\\ &= 2026+\left(\left[ \sum_{i=0}^\infty a_i \right]-a_0-a_1\right)-\frac{S}{8}\\ &= 2026+(S-3-1)-\frac{S}{8}\\ &= 2022+\frac{7S}{8}.\\ \end{align*}$ We have $\begin{align*} S &= 2022+\frac{7S}{8}\implies\\ \frac{S}{8} &= 2022\implies\\ S &= 8\cdot2022 = \boxed{16176}. \end{align*}$

~pinkpig

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2023 SSMO Relay Round 1 Problems/Problem 2

Problem

Solution