Difference between revisions of "2002 AMC 10P Problems/Problem 23"
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\begin{align*} | \begin{align*} | ||
− | + | 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{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}. | + | &=\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}. \\ | ||
\end{align*} | \end{align*} | ||
Revision as of 05:54, 15 July 2024
Problem
Let
and
Find the integer closest to
Solution 1
Start by subtracting and and group those with a common denominator together, leaving and to the side.
\begin{align*} 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}. \\ \end{align*}
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 |
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