1992 AIME Problems/Problem 4

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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 $3: 4: 5$?


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, $n = \boxed{062}$

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