Difference between revisions of "2001 AIME II Problems/Problem 3"
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== Problem == | == Problem == | ||
Given that | Given that | ||
+ | |||
<cmath> | <cmath> | ||
\begin{align*}x_{1}&=211,\\ | \begin{align*}x_{1}&=211,\\ | ||
Line 8: | Line 9: | ||
x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*} | x_{n}&=x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4}\ \text{when}\ n\geq5, \end{align*} | ||
</cmath> | </cmath> | ||
+ | |||
find the value of <math>x_{531}+x_{753}+x_{975}</math>. | find the value of <math>x_{531}+x_{753}+x_{975}</math>. | ||
== Solution == | == Solution == | ||
− | <math>x_5=x_4-x_3+x_2-x_1</math> | + | We find that <math>x_5 = 267</math> by the recursive formula. Summing the [[recursion]]s |
+ | |||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | yields <math>x_{n} = -x_{n-5}</math>. Thus <math>x_n = (-1)^k x_{n-5k}</math>. Since <math>531 = 106 \cdot 5 + 1,\ 753 = 150 \cdot 5 + 3,\ 975 = 194 \cdot 5 + 5</math>, it follows that | ||
+ | |||
+ | <cmath>x_{531} + x_{753} + x_{975} = (-1)^{106} x_1 + (-1)^{150} x_3 + (-1)^{194} x_5 = 211 + 420 + 267 = \boxed{898}.</cmath> | ||
+ | |||
+ | == Solution Variant == | ||
+ | The recursive formula suggests telescoping. Indeed, if we add <math>x_n</math> and <math>x_{n-1}</math>, we have <math>x_n + x_{n-1} = (x_{n-1} - x_{n-2} + x_{n-3} - x_{n-4}) + (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) = x_{n-1} - x_{n-5}</math>. | ||
+ | |||
+ | Subtracting <math>x_{n-1}</math> yields <math>x_n = -x_{n-5} \implies x_n = -(-(x_{n-10})) = x_{n-10}</math>. | ||
+ | |||
+ | Thus, | ||
+ | |||
+ | <cmath>x_{531} + x_{753} + x_{975} = x_1 + x_3 + x_5 = x_1 + x_3 + (x_4 - x_3 + x_2 - x_1) = x_2 + x_4 = 375 + 523 = \boxed{898}.</cmath> | ||
+ | |||
+ | Notice that we didn't need to use the values of <math>x_1</math> or <math>x_3</math> at all. | ||
− | |||
− | + | ==Non-Rigorous Solution== | |
− | + | Calculate the first few terms: | |
− | < | + | <cmath>211,375,420,523,267,-211,-375,-420,-523,\dots</cmath> |
− | + | At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously <math>211+420+267=\boxed{898}</math> | |
− | + | ~Dhillonr25 | |
− | + | ||
− | + | == Video Solution by OmegaLearn == | |
− | + | https://youtu.be/lH-0ul1hwKw?t=870 | |
− | + | ||
+ | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
{{AIME box|year=2001|n=II|num-b=2|num-a=4}} | {{AIME box|year=2001|n=II|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:28, 12 November 2022
Contents
Problem
Given that
find the value of .
Solution
We find that by the recursive formula. Summing the recursions
yields . Thus
. Since
, it follows that
Solution Variant
The recursive formula suggests telescoping. Indeed, if we add and
, we have
.
Subtracting yields
.
Thus,
Notice that we didn't need to use the values of or
at all.
Non-Rigorous Solution
Calculate the first few terms:
At this point it is pretty clear that the sequence is periodic with period 10 (one may prove it quite easily like in solution 1) so our answer is obviously
~Dhillonr25
Video Solution by OmegaLearn
https://youtu.be/lH-0ul1hwKw?t=870
~ pi_is_3.14
See also
2001 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.