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Difference between revisions of "1962 AHSME Problems/Problem 40"

(Created page with "==Problem== The limiting sum of the infinite series, <math>\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots</math> whose <math>n</math>th term is <math>\frac{n}{10^n}</math...")
 
(4 intermediate revisions by 3 users not shown)
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==Solution==
 
==Solution==
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"Unsolved"
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The series can be written as the following:
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<math>\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + ...</math>
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<math>+ \frac{1}{10^2} + \frac{1}{10^3} + \frac{1}{10^4} + ...</math>
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<math>+ \frac{1}{10^3} + \frac{1}{10^4} + \frac{1}{10^5} + ...</math>
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and so on.
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by using the formula for infinite geometric series <math>(\frac{a}{1-r})</math>,
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We can get <math>\frac{\frac{1}{10}}{1-\frac{1}{10}}</math> <math>+</math> <math>\frac{\frac{1}{10^2}}{1-\frac{1}{10}}</math> <math>+</math> <math>\frac{\frac{1}{10^3}}{1-\frac{1}{10}}</math> <math>+</math> ...
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Since they all have common denominators, we get <math>\frac{(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3})}{\frac{9}{10}}</math>.
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Using the infinite series formula again, we get <math>\frac{\frac{\frac{1}{10}}{1-\frac{1}{10}}}{\frac{9}{10}}</math> <math>=</math> <math>\frac{\frac{\frac{1}{10}}{\frac{9}{10}}}{\frac{9}{10}}</math> <math>=</math> <math>\frac{\frac{1}{9}}{\frac{9}{10}}</math> <math>=</math> <math>\boxed{ (B) \frac{10}{81}}</math>
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==Solution 2==
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So.. we have the sum to be <math>\frac{1}{10}+\frac{2}{100}+\frac{3}{1000}</math>...
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Notice that this can be written as
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<math>\frac{1}{10}+\frac{0.2}{10}+\frac{0.03}{10}+\frac{0.004}{10}</math>.
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Now, it is trivial that the new fraction we seek is <math>\frac{1.234567891011......}{10}</math>
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Testing the answer choices, we see that <math>\boxed{B}</math> is the correct answer.
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==Video Solution==
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Problem starts at 2:20 : https://www.youtube.com/watch?v=3PDZtddYQoM&t=5s

Revision as of 02:01, 16 August 2020

Problem

The limiting sum of the infinite series, $\frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + \dots$ whose $n$th term is $\frac{n}{10^n}$ is:

$\textbf{(A)}\ \frac{1}9\qquad\textbf{(B)}\ \frac{10}{81}\qquad\textbf{(C)}\ \frac{1}8\qquad\textbf{(D)}\ \frac{17}{72}\qquad\textbf{(E)}\ \text{larger than any finite quantity}$

Solution

The series can be written as the following:

$\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + ...$

$+ \frac{1}{10^2} + \frac{1}{10^3} + \frac{1}{10^4} + ...$

$+ \frac{1}{10^3} + \frac{1}{10^4} + \frac{1}{10^5} + ...$

and so on.

by using the formula for infinite geometric series $(\frac{a}{1-r})$,

We can get $\frac{\frac{1}{10}}{1-\frac{1}{10}}$ $+$ $\frac{\frac{1}{10^2}}{1-\frac{1}{10}}$ $+$ $\frac{\frac{1}{10^3}}{1-\frac{1}{10}}$ $+$ ... Since they all have common denominators, we get $\frac{(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3})}{\frac{9}{10}}$. Using the infinite series formula again, we get $\frac{\frac{\frac{1}{10}}{1-\frac{1}{10}}}{\frac{9}{10}}$ $=$ $\frac{\frac{\frac{1}{10}}{\frac{9}{10}}}{\frac{9}{10}}$ $=$ $\frac{\frac{1}{9}}{\frac{9}{10}}$ $=$ $\boxed{ (B) \frac{10}{81}}$

Solution 2

So.. we have the sum to be $\frac{1}{10}+\frac{2}{100}+\frac{3}{1000}$... Notice that this can be written as $\frac{1}{10}+\frac{0.2}{10}+\frac{0.03}{10}+\frac{0.004}{10}$. Now, it is trivial that the new fraction we seek is $\frac{1.234567891011......}{10}$

Testing the answer choices, we see that $\boxed{B}$ is the correct answer.

Video Solution

Problem starts at 2:20 : https://www.youtube.com/watch?v=3PDZtddYQoM&t=5s

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