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

Revision as of 20:05, 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 $A_{k}^{}=\frac{N!}{k!(N-k)!}x^{k}=\frac{(N-k+1)!}{k!}x^{k}$ . Taking logarithms in both sides of this last equation, and recalling that $\log(a_{}^{}b)=\log a + \log b$ (valid if $a_{}^{},b_{}^{}>0$ ), we have

$\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)}{j}x\right]$

@@ Line 10: / Line 10: @@
 <math>
-\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]
+\log(A_{k})=\log\left[\frac{(N-k+1)!}{k!}x^{k}\right]=\sum_{j=1}^{k}\log\left[\frac{(N-j+1)}{j}x\right]
 </math>
 == See also ==
 {{AIME box|year=1991|num-b=2|num-a=4}}

1991 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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Difference between revisions of "1991 AIME Problems/Problem 3"

Revision as of 20:05, 20 April 2007

Problem

Solution

See also