Difference between revisions of "2018 AMC 8 Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | By adding up the numbers in each of the 6 | + | By adding up the numbers in each of the 6 parentheses, we have: |
− | + | <math>\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}</math>. | |
Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus the answer would be <math>\boxed{\textbf{(D) }7}</math>. | Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus the answer would be <math>\boxed{\textbf{(D) }7}</math>. |
Revision as of 21:18, 28 October 2019
Problem 2
What is the value of the product
Solution
By adding up the numbers in each of the 6 parentheses, we have:
.
Using telescoping, most of the terms cancel out diagonally. We are left with which is equivalent to
. Thus the answer would be
.
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.