Difference between revisions of "1991 AIME Problems/Problem 3"
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Expanding <math>(1+0.2)^{1000}_{}</math> by the binomial theorem and doing no further manipulation gives | Expanding <math>(1+0.2)^{1000}_{}</math> by the binomial theorem and doing no further manipulation gives | ||
| − | <math>{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}</math | + | <math>{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}</math> |
| − | + | <math>= A_0 + A_1 + A_2 + \cdots + A_{1000},</math> | |
where <math>A_k = {1000 \choose k}(0.2)^k</math> for <math>k = 0,1,2,\ldots,1000</math> | where <math>A_k = {1000 \choose k}(0.2)^k</math> for <math>k = 0,1,2,\ldots,1000</math> | ||
Revision as of 19:00, 20 April 2007
Problem
Expanding 
by the binomial theorem and doing no further manipulation gives
where 
for 
For which 
is 
the largest?
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
