Difference between revisions of "1962 AHSME Problems/Problem 40"
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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>. | 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>. | ||
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> | 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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | So.. we have the sum to be <math>\frac{1}{10}+\frac{2}{100}+\frac{3}{1000}</math>... | ||
+ | Notice that this can be written as | ||
+ | <math>\frac{1}{10}+\frac{0.2}{10}+\frac{0.03}{10}+\frac{0.004}{10}</math>. | ||
+ | Now, it is trivial that the new fraction we seek is <math>\frac{1.234567891011......}{10}</math> | ||
+ | Now, testing the answer choices, we see that <math>\boxed{B}</math> is the correct answer. |
Revision as of 22:02, 24 July 2017
Problem
The limiting sum of the infinite series, whose
th term is
is:
Solution
The series can be written as the following:
and so on.
by using the formula for infinite geometric series ,
We can get
...
Since they all have common denominators, we get
.
Using the infinite series formula again, we get
Solution 2
So.. we have the sum to be ...
Notice that this can be written as
.
Now, it is trivial that the new fraction we seek is
Now, testing the answer choices, we see that
is the correct answer.