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Difference between revisions of "2023 SSMO Relay Round 1 Problems/Problem 2"
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==Problem==
 
==Problem==
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Let <math>T=TNYWR</math>. Let <math>a_0 = 3, a_1 = 1, a_2 = N</math>, and let <math>a_n = a_{n-1} - \frac{a_{n-3}}{8}</math>. Find <cmath>\sum_{i=0}^\infty a_i.</cmath>
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Let <math>T=TNYWR</math>. Let <math>a_0 = 3, a_1 = 1, a_2 = T</math>, and let <math>a_n = a_{n-1} - \frac{a_{n-3}}{8}</math> for <math>n\ge 3.</math> Find <cmath>\sum_{i=0}^\infty a_i.</cmath>
  
 
==Solution==
 
==Solution==
 +
We have <math>T = 2022</math>. Let
 +
<cmath>\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*}</cmath>
 +
We have
 +
<cmath>\begin{align*}
 +
S &= 2022+\frac{7S}{8}\implies\\
 +
\frac{S}{8} &= 2022\implies\\
 +
S &= 8\cdot2022 = \boxed{16176}.
 +
\end{align*}</cmath>
 +
 +
~pinkpig

Latest revision as of 17:15, 15 September 2025

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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