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1968 AHSME Problems/Problem 17: Difference between revisions

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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 19:54, 17 July 2024

Problem

Let $f(n)=\frac{x_1+x_2+\cdots +x_n}{n}$, where $n$ is a positive integer. If $x_k=(-1)^k, k=1,2,\cdots ,n$, the set of possible values of $f(n)$ is:

$\text{(A) } \{0\}\quad \text{(B) } \{\frac{1}{n}\}\quad \text{(C) } \{0,-\frac{1}{n}\}\quad \text{(D) } \{0,\frac{1}{n}\}\quad \text{(E) } \{1,\frac{1}{n}\}$

Solution

Because we start with $x_1=-1$, and the terms $x_k$ alternate between $1$ and $-1$, there is either one more $-1$ than the number of $1$s (when $n$ is odd), or there are an equal number of $1$s and $-1$s (when $n$ is even). In the former case, $f(n)=\frac{-1}{n}$, and, in the latter case, $f(n)=0$. This is only consistent with answer $\fbox{C}$.

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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