Difference between revisions of "1986 AIME Problems/Problem 11"
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We want the coefficient of the <math>y^2</math> term of each power of each binomial, which by the binomial theorem is <math>{2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}</math>. The [[Hockey Stick Identity]] tells us that this quantity is equal to <math>{18\choose 3} = \boxed{816}</math>. | We want the coefficient of the <math>y^2</math> term of each power of each binomial, which by the binomial theorem is <math>{2\choose 2} + {3\choose 2} + \cdots + {17\choose 2}</math>. The [[Hockey Stick Identity]] tells us that this quantity is equal to <math>{18\choose 3} = \boxed{816}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Again, notice <math>x=y-1</math>. Substituting <math>y-1</math> for <math>x</math> in <math>f(x)</math> gives: | ||
+ | <cmath>\begin{align*}1 - x + x^2 + \cdots - x^{17} & = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \cdots - (y - 1)^{17} \\ | ||
+ | & = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.</cmath> | ||
+ | From binomial theorem, the coefficient of the <math>y^2</math> term is <math>{2\choose 0} + {3\choose 1} + \cdots + {17\choose 15}</math>. This is actually the sum of the first 16 triangular numbers, which evaluates to <math>\frac{(16)(17)(18)}{6} = \boxed{816}</math>. | ||
== See also == | == See also == |
Revision as of 00:47, 1 January 2020
Problem
The polynomial may be written in the form
, where
and the
's are constants. Find the value of
.
Solution
Solution 1
Using the geometric series formula, . Since
, this becomes
. We want
, which is the coefficient of the
term in
(because the
in the denominator reduces the degrees in the numerator by
). By the Binomial Theorem, this is
.
Solution 2
Again, notice . So
We want the coefficient of the term of each power of each binomial, which by the binomial theorem is
. The Hockey Stick Identity tells us that this quantity is equal to
.
Solution 3
Again, notice . Substituting
for
in
gives:
From binomial theorem, the coefficient of the
term is
. This is actually the sum of the first 16 triangular numbers, which evaluates to
.
See also
1986 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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