Difference between revisions of "1968 AHSME Problems/Problem 17"
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== Solution == | == Solution == | ||
− | Because we start with <math>x_1=-1</math>, and the terms <math>x_k</math> alternate between <math>1</math> and <math>-1</math>, there is either one more <math>-1</math> than the number of <math>1</math>s (when <math>n</math> is odd), or there are an equal number of <math>1</math>s and <math>-1</math>s. In the former case, <math>f(n)=\frac{-1}{n}</math>, and, in the latter case, <math>f(n)=0</math>. This is only consistent with answer <math>\fbox{C}</math>. | + | Because we start with <math>x_1=-1</math>, and the terms <math>x_k</math> alternate between <math>1</math> and <math>-1</math>, there is either one more <math>-1</math> than the number of <math>1</math>s (when <math>n</math> is odd), or there are an equal number of <math>1</math>s and <math>-1</math>s (when <math>n</math> is even). In the former case, <math>f(n)=\frac{-1}{n}</math>, and, in the latter case, <math>f(n)=0</math>. This is only consistent with answer <math>\fbox{C}</math>. |
== See also == | == See also == |
Latest revision as of 20:54, 17 July 2024
Problem
Let , where
is a positive integer. If
, the set of possible values of
is:
Solution
Because we start with , and the terms
alternate between
and
, there is either one more
than the number of
s (when
is odd), or there are an equal number of
s and
s (when
is even). In the former case,
, and, in the latter case,
. This is only consistent with answer
.
See also
1968 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
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