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| ==Solution== | | ==Solution== |
− | We have <math>T = 2022</math>. Let
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− | <cmath>\begin{align*}
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− | S = \sum_{i=0}^\infty a_i &= a_0+a_1+a_2+\sum_{i=3}^\infty a_i\\
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− | &= 0+1+2022+\sum_{i=3}^\infty \left(a_{i-1}-\frac{a_{i-3}}{8}\right)\\
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− | &= 2023+\sum_{i=3}^{\infty}a_{i-1}-\sum_{i=3}^\infty\frac{a_{i-3}}{8}\\
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− | &= 2023+\sum_{i=2}^\infty a_i-\frac{\sum_{i=0}a_i}{8}\\
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− | &= 2023+\left(\sum_{i=0}^\infty a_i-a_0-a_1\right)-\frac{S}{8}\\
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− | &= 2023+(S-0-1)-\frac{S}{8}\\
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− | &= 2022+\frac{7S}{8}.\\
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− | \end{align*}</cmath>
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− | We have
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− | <cmath>\begin{align*}
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− | S &= 2022+\frac{7S}{8}\implies\\
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− | \frac{S}{8} &= 2022\implies\\
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− | S &= 8\cdot2022 = \boxed{16176}.
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− | \end{align*}</cmath>
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− | ~pinkpig
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Latest revision as of 11:23, 15 September 2025