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Revision as of 15:09, 26 November 2019
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:
![$\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8$](http://latex.artofproblemsolving.com/6/c/9/6c95ac84579064565eceaf3f568f3b6d602a7e33.png)
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
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
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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