Difference between revisions of "2002 AMC 10P Problems/Problem 23"
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</math> | </math> | ||
| − | == Solution 1== | + | == Solution 1 == |
| − | + | Start by subtracting <math>a</math> and <math>b</math> and group those with a common denominator together, leaving <math>\frac{1^2}{1}</math> and <math>\frac{1001^2}{2003}</math> to the side. | |
| + | <math>a-b=\frac{1^2}{1} + \frac{2^2}{3} + \frac{3^2}{5} + \; \dots \; + \frac{1001^2}{2001} | ||
| + | - \frac{1^2}{3} + \frac{2^2}{5} + \frac{3^2}{7} + \; \dots \; + \frac{1001^2}{2003} | ||
| + | =\frac{(2^2-1^2)}{3}+\frac{(3^2-2^2)}{5}+\frac{4^2-3^2}{7} + \; \dots \; + \frac{1001^2-1000^2}{2001} + 1 - \frac{1001^2}{2003}.</math> | ||
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|num-b=22|num-a=24}} | {{AMC10 box|year=2002|ab=P|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 05:52, 15 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
![\[b=\frac{1^2}{3} + \frac{2^2}{5} + \frac{3^2}{7} + \; \dots \; + \frac{1001^2}{2003}.\]](http://latex.artofproblemsolving.com/f/8/6/f86575c256e324ddd1158f0937d2855e81aad62d.png)
Find the integer closest to 
Solution 1
Start by subtracting 
and 
and group those with a common denominator together, leaving 
and 
to the side.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
