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

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== Solution ==
 
== Solution ==
  
In Pascal's Triangle, we know that the binomial coefficients of the nth row are <math>\displaystyle n\chose 0</math>, <math>\displaystyle \binom{n}{1}</math>,...,<math>\displaystyle \binom{n}{n}</math>. Let our row be the nth row such that the three consecutive entries are <math>\displaystyle \binom{n}{r}</math>, <math>\displaystyle \binom{n}{r+1}</math>, and <math>\displaystyle \binom{n}{r+2}</math>.  
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In Pascal's Triangle, we know that the binomial coefficients of the nth row are <math>\displaystyle\binom{n}{0}</math>, <math>\displaystyle \binom{n}{1}</math>,...,<math>\displaystyle \binom{n}{n}</math>. Let our row be the nth row such that the three consecutive entries are <math>\displaystyle \binom{n}{r}</math>, <math>\displaystyle \binom{n}{r+1}</math>, and <math>\displaystyle \binom{n}{r+2}</math>.  
  
 
After expanding and dividing one entry by another (to clean up the factorials), we see that <math>\displaystyle \frac 34=\frac{r+1}{n-r}</math> and <math>\displaystyle \frac45=\frac{r+2}{n-r-1}</math>. Solving, <math>\displaystyle n = \boxed{062}</math>
 
After expanding and dividing one entry by another (to clean up the factorials), we see that <math>\displaystyle \frac 34=\frac{r+1}{n-r}</math> and <math>\displaystyle \frac45=\frac{r+2}{n-r-1}</math>. Solving, <math>\displaystyle n = \boxed{062}</math>

Revision as of 19:18, 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 $\displaystyle\binom{n}{0}$, $\displaystyle \binom{n}{1}$,...,$\displaystyle \binom{n}{n}$. Let our row be the nth row such that the three consecutive entries are $\displaystyle \binom{n}{r}$, $\displaystyle \binom{n}{r+1}$, and $\displaystyle \binom{n}{r+2}$.

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

See also

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