Difference between revisions of "1991 AIME Problems/Problem 3"

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
Let <math>0<x<1</math>. Then we may write <math>A_{k}^{}=\frac{N!}{k!(N-k)!}x^{k}</math>.
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Let <math>0<x_{}^{}<1</math>. Then we may write <math>A_{k}^{}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=2|num-a=4}}
 
{{AIME box|year=1991|num-b=2|num-a=4}}

Revision as of 20:55, 20 April 2007

Problem

Expanding $(1+0.2)^{1000}_{}$ by the binomial theorem and doing no further manipulation gives

${1000 \choose 0}(0.2)^0+{1000 \choose 1}(0.2)^1+{1000 \choose 2}(0.2)^2+\cdots+{1000 \choose 1000}(0.2)^{1000}$ $= A_0 + A_1 + A_2 + \cdots + A_{1000},$ where $A_k = {1000 \choose k}(0.2)^k$ for $k = 0,1,2,\ldots,1000$. For which $k_{}^{}$ is $A_k^{}$ the largest?

Solution

Let $0<x_{}^{}<1$. Then we may write $A_{k}^{}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}$.

See also

1991 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions