Difference between revisions of "2004 AIME I Problems/Problem 7"
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=== Solution 2 === | === Solution 2 === | ||
Let <math>S</math> be the [[set]] of integers <math>\{-1,2,-3,\ldots,14,-15\}</math>. The coefficient of <math>x^2</math> in the expansion is equal to the sum of the product of each pair of distinct terms, or <math>C = \sum_{1 \le i \neq j}^{15} S_iS_j</math>. Also, we know that | Let <math>S</math> be the [[set]] of integers <math>\{-1,2,-3,\ldots,14,-15\}</math>. The coefficient of <math>x^2</math> in the expansion is equal to the sum of the product of each pair of distinct terms, or <math>C = \sum_{1 \le i \neq j}^{15} S_iS_j</math>. Also, we know that | ||
− | < | + | <cmath>\begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}</cmath> |
where the left-hand sum can be computed from: | where the left-hand sum can be computed from: | ||
<center><math>\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8</math></center> | <center><math>\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8</math></center> |
Revision as of 16:48, 13 March 2015
Problem
Let be the coefficient of in the expansion of the product Find
Solution
Solution 1
Let our polynomial be .
It is clear that the coefficient of in is , so , where is some polynomial divisible by .
Then and so , where is some polynomial divisible by .
However, we also know .
Equating coefficients, we have , so and .
Solution 2
Let be the set of integers . The coefficient of in the expansion is equal to the sum of the product of each pair of distinct terms, or . Also, we know that where the left-hand sum can be computed from:
and the right-hand sum comes from the formula for the sum of the first perfect squares. Therefore, .
See also
2004 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.