Difference between revisions of "1992 AIME Problems/Problem 4"

 
Line 1: Line 1:
 
== Problem ==
 
== 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 <math>3: 4: 5</math>?
  
 
== Solution ==
 
== Solution ==
 +
 +
In Pascal's Triangle, we know that the binomial coefficients of the nth row are <math>\binom{n}{0}</math>, <math>\binom{n}{1}</math>,...,<math>\binom{n}{n}</math>. Let our row be the nth row such that the three consecutive entries are <math>\binom{n}{r}</math>, <math>\binom{n}{r+1}</math>, and <math>\binom{n}{r+2}</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 = \boxed{062}</math>
 +
  
 
== See also ==
 
== See also ==
 
* [[1992 AIME Problems]]
 
* [[1992 AIME Problems]]

Revision as of 19:15, 4 November 2006

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 $3: 4: 5$?

Solution

In Pascal's Triangle, we know that the binomial coefficients of the nth row are $\binom{n}{0}$, $\binom{n}{1}$,...,$\binom{n}{n}$. Let our row be the nth row such that the three consecutive entries are $\binom{n}{r}$, $\binom{n}{r+1}$, and $\binom{n}{r+2}$.

After expanding and dividing one entry by another (to clean up the factorials), we see that $\frac 34=\frac{r+1}{n-r}$ and $\frac45=\frac{r+2}{n-r-1}$. Solving, $n = \boxed{062}$


See also

Invalid username
Login to AoPS