Difference between revisions of "1968 AHSME Problems/Problem 17"
m (→See also) |
m (added "when n is even" for consistency) |
||
| (One intermediate revision by the same user not shown) | |||
| Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
| − | <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 19:54, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
