Difference between revisions of "2022 AMC 12A Problems/Problem 8"
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By continuing this, we get the form | By continuing this, we get the form | ||
− | <math>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} ...</math> | + | <math>10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdot ...</math> |
which is | which is | ||
− | <math>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ...}</math>. | + | <math>10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + ...}</math>. |
Using the formula for an infinite geometric series <math>S = \frac{a}{1-r}</math>, we get | Using the formula for an infinite geometric series <math>S = \frac{a}{1-r}</math>, we get | ||
− | <math>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} ... = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}</math> | + | <math>\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + ... = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}</math> |
Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}</math>. | Thus, our answer is <math>10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}</math>. |
Revision as of 16:42, 15 October 2023
Contents
[hide]Problem
The infinite product evaluates to a real number. What is that number?
Solution 1
We can write as . Similarly, .
By continuing this, we get the form
which is
.
Using the formula for an infinite geometric series , we get
Thus, our answer is .
- phuang1024
Solution 2
We can write this infinite product as (we know from the answer choices that the product must converge):
If we raise everything to the power, we get:
Since is positive (it is an infinite product of positive numbers), it must be that .
~ Oxymoronic15
Solution 3
Move the first term inside the second radical. We get
Do this for the third radical as well.
It is clear what the pattern is. Setting the answer as we have
~kxiang
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.