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2001 AIME II Problems/Problem 3

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Problem

Given that

\begin{align*}x_{1}&=211,\\x_{2}&=375,\\x_{3}&=420,\\x_{4}&=523,\ \text{and}\\x_{n}&=x_{n-1}-x_{n-2}+x_{n...

find the value of x_{531}+x_{753}+x_{975}.

Solution

We find that x_5 = 267 by the recursive formula. Summing the recursions

\begin{align*}x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \\x_{n-1}&=x_{n-2}-x_{n-3}+x_{n-4}-x_{n-5}\end{align*}

yields x_{n} = -x_{n-5}. Thus x_n = (-1)^k x_{n-5k}. Since 531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5, it follows that

x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.

See also

2001 AIME II (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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