Difference between revisions of "2002 AMC 10P Problems/Problem 23"
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| + | == Problem == | ||
| + | |||
| + | Let | ||
| + | <cmath>a=\frac{1^2}{1} + \frac{2^2}{3} + \frac{3^2}{5} + \; \dots \; + \frac{1001^2}{2001}</cmath> | ||
| + | |||
| + | and | ||
| + | |||
| + | <cmath>a=\frac{1^2}{3} + \frac{2^2}{5} + \frac{3^2}{7} + \; \dots \; + \frac{1001^2}{2003}.</cmath> | ||
| + | |||
| + | Find the integer closest to <math>a-b.</math> | ||
| + | |||
| + | <math> | ||
| + | \text{(A) }500 | ||
| + | \qquad | ||
| + | \text{(B) }501 | ||
| + | \qquad | ||
| + | \text{(C) }999 | ||
| + | \qquad | ||
| + | \text{(D) }1000 | ||
| + | \qquad | ||
| + | \text{(E) }1001 | ||
| + | </math> | ||
| + | |||
== Solution 1== | == Solution 1== | ||
Revision as of 18:01, 14 July 2024
Problem
Let
![\[a=\frac{1^2}{1} + \frac{2^2}{3} + \frac{3^2}{5} + \; \dots \; + \frac{1001^2}{2001}\]](http://latex.artofproblemsolving.com/8/1/4/81414ced90ee5723c960e8677bf115e72410e4bb.png)
and
![\[a=\frac{1^2}{3} + \frac{2^2}{5} + \frac{3^2}{7} + \; \dots \; + \frac{1001^2}{2003}.\]](http://latex.artofproblemsolving.com/5/e/7/5e705381a9cdde1f763caee336397ba2f8429d3c.png)
Find the integer closest to 
Solution 1
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
