2004 AIME I Problems/Problem 7
Problem
Let be the coefficient of in the expansion of the product Find
Contents
Solutions
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, .
Solution 3 (Bash)
Consider the set . Denote by all size 2 subsets of this set. Replace each element of by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to or , we can simplify this to
Solution 4 (Easy)
We know that this polynomial has roots and the coefficient of will be the sum of the product taken by . However, since this is closer to the constant side, we can create a new polynomial with the reciprocal roots which will make the coefficient of and thus the sum of the reciprocal roots taken by 2. We can calculate this with . This will give us giving us a final answer of .
~ Vedoral
Solution 4
Let set be and set be . The sum of the negative coefficients is the sum of the products of the elements in all two element sets such that one element is from and the other is from . Each summand is a term in the expansion of which equals . The sum of the positive coefficients is the sum of the products of all two element sets such that the two elements are either both in or both in . By counting, the sum is , so the sum of all coefficients is . Thus, the answer is .
Solution 5
We can find out the coefficient of by multiplying every pair of two coefficients for . This means that we multiply by and by . and etc. This sum can be easily simplified and is equal to or .
-David Camacho
Solution 6
This is just another way of summing the subsets of 2 from . Start from the right and multiply -15 to everything on its left. Use the distributive property and add all the 14 integers together to get 7. This gives us . Doing this for 14 gives us , and for -13 we get . This pattern repeats where every two integers will multiple 7, 6,... to 0. Combining and simplifying the pattern give us this: . The expression gives us -588, or . This is a good solution because it guarantees we never add a product twice, and the pattern is simple to add by hand.
-jackshi2006
Solution 7
We expand and obtain
Do not do this in an actual competition.
~Sliced_Bread
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 |
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