Difference between revisions of "1986 AIME Problems/Problem 11"
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& = 1 + (1 - y) + (1 - y)^2 + (1 - y)^3 \cdots + (1 - y)^{17}\end{align*}.</cmath> | & = 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>. | 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>. | ||
+ | |||
+ | == Solution 4(calculus) == | ||
+ | Let f(x) be | ||
== See also == | == See also == |
Revision as of 23:45, 3 March 2020
Problem
The polynomial may be written in the form , where and the 's are constants. Find the value of .
Contents
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 .
Solution 4(calculus)
Let f(x) be
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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