Difference between revisions of "2018 AMC 8 Problems/Problem 2"
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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>. | ||
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==See Also== | ==See Also== | ||
Revision as of 15:02, 11 June 2020
Problem 2
What is the value of the product![\[\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?\]](http://latex.artofproblemsolving.com/2/c/7/2c7aacf4c691c580906f1aaa847be38a7dbf6f26.png)
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. 
