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2016 AMC 12B Problems/Problem 14

Revision as of 23:37, 5 August 2017 by Obtuse (talk | contribs) (→‎Solution 2)

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

A geometric sequence always looks like

\[a,ar,ar^2,ar^3,\dots\]

and they say that the second term $ar=1$. You should know that the sum of an infinite geometric series (denoted by $S$ here) is $\frac{a}{1-r}$. We now have a system of equations which allows us to find $S$ in one variable.

\begin{align*} ar&=1 \\ S&=\frac{a}{1-r} \end{align*}

$\textbf{Solving in terms of \textit{a} then graphing}$

\[S=\frac{a^2}{a-1}\]

We seek the smallest positive value of $S$. We proceed by graphing in the $aS$ plane and find the answer is $\boxed{\textbf{(E)}\ 4}.$

$\textbf{Solving in terms of \textit{r} then graphing}$

\[S=\frac{1}{-r^2+r}\]

We seek the smallest positive value of $S$. We proceed by graphing in the $rS$ plane and find the answer is $\boxed{\textbf{(E)}\ 4}.$

$\textbf{Solving in terms of \textit{a} then doing some calculus}$

\[S=\frac{a^2}{a-1}\]

We seek the smallest positive value of $S$. $\frac{(a-2)a}{(a-1)^2}=S'$ and $\frac{(a-2)a}{(a-1)^2}=0$ at $a=0$ and $a=2$. $\frac{2}{(a-1)^3}=S''$ and $\frac{2}{(0-1)^3}$ is negative (implying a relative maximum occurs at $a=0$) and $\frac{2}{(2-1)^3}$ is positive (implying a relative minimum occurs at $a=2$). At $a=2$, $S=4$. Since this a competition math problem with an answer, we know this relative minimum must also be the absolute minimum among the "positive parts" of $S$ and that our answer is indeed $\boxed{\textbf{(E)}\ 4}.$ However, to be sure of this outside of this cop-out, one can analyze the end behavior of $S$, how $S$ behaves at its asymptotes, and the locations of its maxima and minima relative to the asymptotes to be sure that 4 is the absolute minimum among the "positive parts" of $S$.

$\textbf{Solving in terms of \textit{r} then being clever}$

\[S=\frac{1}{-r^2+r}\]

We seek the smallest positive value of $S$. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at $x=\frac{1}{2}$ and $\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}$.

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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