Difference between revisions of "1962 AHSME Problems/Problem 40"
(→Solution 2) |
|||
(One intermediate revision by one other user not shown) | |||
Line 28: | Line 28: | ||
Testing the answer choices, we see that <math>\boxed{B}</math> is the correct answer. | Testing the answer choices, we see that <math>\boxed{B}</math> is the correct answer. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let | ||
+ | <cmath> | ||
+ | S = \frac{1}{10} + \frac{2}{10^2} + \frac{3}{10^3} + ... | ||
+ | </cmath> | ||
+ | Then | ||
+ | <cmath> | ||
+ | 10S = 1 + \frac{2}{10} + \frac{3}{10^2} + \frac{4}{10^3} + ... | ||
+ | </cmath> | ||
+ | Subtracting <math>1S</math> from <math>10S</math>, we got: | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 9S &= 1 + \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + ... \\ | ||
+ | &= \frac{1}{1-\frac{1}{10}} = \frac{10}{9} \\ | ||
+ | S &= \frac{10}{81} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | Therefore, the answer is <math>\boxed{(B) \frac{10}{81}}</math>. -nullptr07 | ||
+ | |||
+ | ==Video Solution== | ||
+ | Problem starts at 2:20 : https://www.youtube.com/watch?v=3PDZtddYQoM&t=5s |
Latest revision as of 16:01, 28 June 2023
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
Testing the answer choices, we see that is the correct answer.
Solution 3
Let Then Subtracting from , we got: Therefore, the answer is . -nullptr07
Video Solution
Problem starts at 2:20 : https://www.youtube.com/watch?v=3PDZtddYQoM&t=5s