Difference between revisions of "1962 AHSME Problems/Problem 40"
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| − | {{ | + | The series can be written as the following: |
| + | |||
| + | <math>\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + ...</math> | ||
| + | |||
| + | <math>+ \frac{1}{10^2} + \frac{1}{10^3} + \frac{1}{10^4} + ...</math> | ||
| + | |||
| + | <math>+ \frac{1}{10^3} + \frac{1}{10^4} + \frac{1}{10^5} + ...</math> | ||
| + | |||
| + | and so on. | ||
| + | |||
| + | by using the formula for infinite geometric series <math>(\frac{a}{1-r})</math>, | ||
| + | |||
| + | 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> ... | ||
| + | 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> | ||
Revision as of 19:27, 19 February 2016
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 
