2006 AIME A Problems/Problem 15
Problem
Given that a sequence satisfies and
for all integers
find the minimum possible value of
Solution
If you take the recursion formula and square both sides, you get , or
. If you take the sum of this from 1 to 2007, you see that
x_{2007}^2-x_0^2=6(x_0+x_1+x_2+\cdots +x_{2006})+9*2007\\ x_{2007}^2=6(S)+9*2007\\
S=\frac{x_{2007}^2-9*2007}{6}\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)where .
Note that x_{2007} is a multiple of 3, and the closest square multiple of 3 to 2*2007 is . Therefore the smallest value of the absolute value of
is
.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Final Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |