Difference between revisions of "1992 AIME Problems/Problem 4"
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In Pascal's Triangle, we know that the binomial coefficients of the <math>n</math>th row are <math>n\choose 0</math>, <math>n\choose1</math>, ..., <math>n\choose n</math>. Let our row be the <math>n</math>th row such that the three consecutive entries are <math>n \choose r</math>, <math>n \choose r+1</math> and <math>n \choose r+2</math>. | In Pascal's Triangle, we know that the binomial coefficients of the <math>n</math>th row are <math>n\choose 0</math>, <math>n\choose1</math>, ..., <math>n\choose n</math>. Let our row be the <math>n</math>th row such that the three consecutive entries are <math>n \choose r</math>, <math>n \choose r+1</math> and <math>n \choose r+2</math>. | ||
| − | After expanding and dividing one entry by another (to clean up the factorials), we see that <math> | + | After expanding and dividing one entry by another (to clean up the factorials), we see that <math>\frac 34=\frac{r+1}{n-r}</math> and <math>\frac45=\frac{r+2}{n-r-1}</math>. Solving, <math>n = 62</math>. |
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 22:57, 11 November 2007
Problem
In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio 
?
Solution
In Pascal's Triangle, we know that the binomial coefficients of the 
th row are 
, 
, ..., 
. Let our row be the th row such that the three consecutive entries are 
, 
and 
.
After expanding and dividing one entry by another (to clean up the factorials), we see that 
and 
. Solving, 
.
| 1992 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
