# Three Consecutive Binomial Coefficients in AP

I came across an interesting pattern in the Pascal triangle, and thought I would post it as a problem here.

Given three consecutive binomial coefficients $$\binom n{r-1},\binom nr,\binom n{r+1}$$ which are in AP, where \begin{align} \binom n{r-1}&=1\cdot 10^m+1\\ \binom nr &=2\cdot 10^m+2\\ \binom n{r+1} &=3\cdot 10^m+3 \end{align} where $m, n, r$ are integers, find $n, r$.

• Hint: $$\binom n{r-1} = \frac{1}{2}\binom n{r} = \frac{1}{3}\binom n{r+1} = 10^m+1$$ – Jose Arnaldo Bebita-Dris Nov 12 '14 at 11:40

It is well known (to those who know such things well) that $1001,2002,3003$ occur consecutively in the $14$th row of Pascal's triangle.
• Thank you for sharing your observation. You certainly know the Pascal triangle very well! I was hoping that someone on MSE might attempt a derivation based on the given problem statement, and in the process also identify if a solution exists for other values of $m$. Have upvoted nevertheless. – hypergeometric Nov 12 '14 at 14:34