Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 12B Problems/Problem 14"

(→‎Solution)
Line 12: Line 12:
 
The second term in a geometric series is <math>a_2 = a \cdot r</math>, where <math>r</math> is the common ratio for the series and <math>a</math> is the first term of the series. So we know that <math>a\cdot r = 1</math> and we wish to find the minimum value of the infinite sum of  the series. We know that: <math>S_\infty = \frac{a}{1-r}</math> and substituting in <math>a=\frac{1}{r}</math>, we get that <math>S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}</math>. From here, you can either use calculus or AM-GM.
 
The second term in a geometric series is <math>a_2 = a \cdot r</math>, where <math>r</math> is the common ratio for the series and <math>a</math> is the first term of the series. So we know that <math>a\cdot r = 1</math> and we wish to find the minimum value of the infinite sum of  the series. We know that: <math>S_\infty = \frac{a}{1-r}</math> and substituting in <math>a=\frac{1}{r}</math>, we get that <math>S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}</math>. From here, you can either use calculus or AM-GM.
  
−
\textbf{Calculus}
+
<math>\textbf{Calculus}</math>
 
Let <math>f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}</math>, then <math>f'(x) = -(x-x^2)^{-2}\cdot (1-2x)</math>. Since <math>f(0)</math> and <math>f(1)</math> are undefined <math>x \neq 0,1</math>. This means that we only need to find where the derivative equals <math>0</math>, meaning <math>1-2x = 0 \Rightarrow x =\frac{1}{2}</math>. So <math> r = \frac{1}{2}</math>, meaning that <math>S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4</math>
 
Let <math>f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}</math>, then <math>f'(x) = -(x-x^2)^{-2}\cdot (1-2x)</math>. Since <math>f(0)</math> and <math>f(1)</math> are undefined <math>x \neq 0,1</math>. This means that we only need to find where the derivative equals <math>0</math>, meaning <math>1-2x = 0 \Rightarrow x =\frac{1}{2}</math>. So <math> r = \frac{1}{2}</math>, meaning that <math>S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4</math>
  
−
\textbf{AM-GM}
+
<math>\textbf{AM-GM}</math>
 
For 2 positive real numbers <math>a</math> and <math>b</math>, <math>\frac{a+b}{2} \geq \sqrt{ab}</math>. Let <math>a = \frac{1}{r}</math> and <math>b = \frac{1}{1-r}</math>. Then: <math>\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}</math>. This implies that <math>\frac{S_\infty}{2} \geq \sqrt{S_\infty}</math>. or <math>S_\infty^2 \geq 4 \cdot S_\infty</math>. Rearranging : <math>(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4</math>. Thus, the smallest value is <math>S_\infty = 4</math>.
 
For 2 positive real numbers <math>a</math> and <math>b</math>, <math>\frac{a+b}{2} \geq \sqrt{ab}</math>. Let <math>a = \frac{1}{r}</math> and <math>b = \frac{1}{1-r}</math>. Then: <math>\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}</math>. This implies that <math>\frac{S_\infty}{2} \geq \sqrt{S_\infty}</math>. or <math>S_\infty^2 \geq 4 \cdot S_\infty</math>. Rearranging : <math>(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4</math>. Thus, the smallest value is <math>S_\infty = 4</math>.
  

Revision as of 12:39, 5 August 2017

Problem

The sum of an infinite geometric series is a positive number $S$, and the second term in the series is $1$. What is the smallest possible value of $S?$

$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution

The second term in a geometric series is $a_2 = a \cdot r$, where $r$ is the common ratio for the series and $a$ is the first term of the series. So we know that $a\cdot r = 1$ and we wish to find the minimum value of the infinite sum of the series. We know that: $S_\infty = \frac{a}{1-r}$ and substituting in $a=\frac{1}{r}$, we get that $S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}$. From here, you can either use calculus or AM-GM.

$\textbf{Calculus}$ Let $f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}$, then $f'(x) = -(x-x^2)^{-2}\cdot (1-2x)$. Since $f(0)$ and $f(1)$ are undefined $x \neq 0,1$. This means that we only need to find where the derivative equals $0$, meaning $1-2x = 0 \Rightarrow x =\frac{1}{2}$. So $r = \frac{1}{2}$, meaning that $S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4$

$\textbf{AM-GM}$ For 2 positive real numbers $a$ and $b$, $\frac{a+b}{2} \geq \sqrt{ab}$. Let $a = \frac{1}{r}$ and $b = \frac{1}{1-r}$. Then: $\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}}=\sqrt{\frac{1}{r}+\frac{1}{1-r}}$. This implies that $\frac{S_\infty}{2} \geq \sqrt{S_\infty}$. or $S_\infty^2 \geq 4 \cdot S_\infty$. Rearranging : $(S_\infty-2)^2 \geq 4 \Rightarrow S_\infty - 2 \geq 2 \Rightarrow S_\infty \geq 4$. Thus, the smallest value is $S_\infty = 4$.

Solution 2

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_14&oldid=86823"