Difference between revisions of "2016 AMC 12B Problems/Problem 14"
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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 13:39, 5 August 2017
Contents
Problem
The sum of an infinite geometric series is a positive number , and the second term in the series is . What is the smallest possible value of
Solution
The second term in a geometric series is , where is the common ratio for the series and is the first term of the series. So we know that and we wish to find the minimum value of the infinite sum of the series. We know that: and substituting in , we get that . From here, you can either use calculus or AM-GM.
Let , then . Since and are undefined . This means that we only need to find where the derivative equals , meaning . So , meaning that
For 2 positive real numbers and , . Let and . Then: . This implies that . or . Rearranging : . Thus, the smallest value is .
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 |
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