Recursive Sequence Value 3
The sequence $ a_0 $, $ a_1 $, $ a_2 $, $ \ldots\, $ satisfies the recurrence equation
\[
a_n = 2 a_{n-1} - 2 a_{n - 2} + a_{n - 3}
\]for every integer $ n \ge 3 $. If $ a_{20} = 1 $, $ a_{25} = 10 $, and $ a_{30} = 100 $, what is the value of $ a_{1331} $?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
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- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
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- $,$
- $\infty$
Solution
We can calculate the first few terms of the sequence and look for a pattern. For $ n=3 $,
$$a_3 = 2a_2 - 2a_1 + a_0.$$For $ n=4 $ we get
$$a_4 = 2a_3 - 2a_2 + a_1 = 2(2a_2 - 2a_1 + a_0) - 2a_2+a_1 = 2a_2 - 3a_1+2a_0.$$With $ n=5 $ we have
$$a_5 = 2a_4 - 2a_3 + a_2 = 2(2a_2 - 3a_1+2a_0) - 2(2a_2 - 2a_1 + a_0) +a_2 = a_2 - 2a_1+2a_0.$$With $ n=6 $ we have
$$a_6 = 2a_5 - 2a_4 + a_3 = 2(a_2 - 2a_1+2a_0) - 2(2a_2 - 3a_1+2a_0)+ 2(2a_2 - 2a_1 + a_0) = a_0.$$Brilliant! We found that $ a_6 = a_0 $ and we can similarly check that $ a_7 = a_1 $ and $ a_8 = a_2 $ and so on because of the recursive rules of the sequence. This means that the sequence is periodic with a period of 6.
This means that $ a_0 = a_{30} = 100 $. Similarly, $ a_1 = a_{25} = 10 $ and $ a_2 = a_{20} = 1 $. Then,
\[a_{1331} = a_5 = a_2 - 2a_1+2a_0 = 1 - 2(10) + 2(100) = \boxed{181}.\]
Freeflows