Difference between revisions of "1959 AHSME Problems/Problem 49"
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Notice that we can arrange the sequence like so: <cmath>\begin{align*}1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots &= 1-\frac{5}{8}-\frac{5}{64}-\frac{5}{512}-\dots \\ &= 1-\left(\sum_{i=1}^{\infty}\frac{5}{8^n}\right) \\ &= 1-\frac{5}{7} \text{ (by the convergence of a geometric series)} \\ &=\frac{2}{7}\end{align*}</cmath> | Notice that we can arrange the sequence like so: <cmath>\begin{align*}1-\frac12-\frac14+\frac18-\frac{1}{16}-\frac{1}{32}+\frac{1}{64}-\frac{1}{128}-\cdots &= 1-\frac{5}{8}-\frac{5}{64}-\frac{5}{512}-\dots \\ &= 1-\left(\sum_{i=1}^{\infty}\frac{5}{8^n}\right) \\ &= 1-\frac{5}{7} \text{ (by the convergence of a geometric series)} \\ &=\frac{2}{7}\end{align*}</cmath> | ||
hence our answer is <math>\fbox{B}</math> | hence our answer is <math>\fbox{B}</math> | ||
| + | |||
| + | == See also == | ||
| + | |||
| + | {{AHSME 50p box|year = 1959|num-b = 48|num-a = 50}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category: AHSME]][[Category:AHSME Problems]] | ||
Latest revision as of 12:06, 22 July 2024
Problem 49
For the infinite series 
let 
be the (limiting) sum. Then equals:
Solution
Notice that we can arrange the sequence like so: 
hence our answer is 
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 48 |
Followed by Problem 50 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
