Difference between revisions of "2004 AIME I Problems/Problem 7"
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<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> | ||
and the right-hand sum comes from the formula for the sum of the first <math>n</math> perfect squares. Therefore, <math>|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}</math>. | and the right-hand sum comes from the formula for the sum of the first <math>n</math> perfect squares. Therefore, <math>|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}</math>. | ||
+ | |||
+ | === Solution 3 (Bash)=== | ||
+ | |||
+ | Consider the set <math>[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]</math> | ||
== See also == | == See also == |
Revision as of 14:27, 24 August 2017
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,
.
Solution 3 (Bash)
Consider the set
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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