Question

Suppose that and are integers with$\mathbf{1}\mathbf{\le }\mathit{k}\mathbf{<}\mathit{n}$ . Prove the hexagon identity$\left(\begin{array}{l}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}n\\ k+1\end{array}\right)\left(\begin{array}{c}n+1\\ k\end{array}\right)\mathbf{=}\left(\begin{array}{c}n-1\\ k\end{array}\right)\left(\begin{array}{c}n\\ k-1\end{array}\right)\left(\begin{array}{l}n+1\\ k+1\end{array}\right)$ which relates terms in Pascal's triangle that form a hexagon.

Short answer

The hexagon identity$\left(\begin{array}{l}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}n\\ k+1\end{array}\right)\left(\begin{array}{c}n+1\\ k\end{array}\right)=\left(\begin{array}{c}n-1\\ k\end{array}\right)\left(\begin{array}{c}n\\ k-1\end{array}\right)\left(\begin{array}{l}n+1\\ k+1\end{array}\right)$ which relates terms in Pascal’s triangle that form a hexagon is proved.

Step-by-step solution

The hexagonal property of Pascal’s triangle

Any hexagon in Pascal's triangle, whose vertices arebinomial coefficients surrounding any entry, has the property that:

• the product of non-adjacent vertices is constant.
• the greatest common divisor of non-adjacent vertices is constant.

Prove the hexagon identity with related terms in Pascal’s triangle

Let n be a positive integer and k an integer with$0\le k\le n$ .

$\begin{array}{r}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}n\\ k+1\end{array}\right)\left(\begin{array}{c}n+1\\ k\end{array}\right)=\frac{(n-1)!}{(k-1)!(n-1-(k-1\left)\right)!}\cdot \frac{n!}{(k+1)!(n-(k+1\left)\right)!}\cdot \frac{(n+1)!}{k!(n+1-k)!}\\ =\frac{(n-1)!}{(k-1)!(n-k)!}\cdot \frac{n!}{(k+1)!(n-k-1))!}\cdot \frac{(n+1)!}{k!(n-k+1)!}\\ =\frac{(n-1)!}{(k-1)!(n-k)!}\cdot \frac{n!}{(k+1)!\left(\right(n-1)-k)!}\cdot \frac{(n+1)!}{k!(n-(k-1\left)\right)!}\end{array}$

(Reorder factors $k!,(k-1)!\text{and}(k+1)!\text{in}$denominators)

$\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}n\\ k+1\end{array}\right)\left(\begin{array}{c}n+1\\ k\end{array}\right)=\frac{(n-1)!}{k!(n-k)!}\cdot \frac{n!}{(k-1)!\left(\right(n-1)-k)!}\cdot \frac{(n+1)!}{(k+1)!(n-(k-1\left)\right)!}$

(Reorder factors $(n-k)!,\left(\right(n-1)-k)!\text{and}(n-(k-1\left)\right)!$denominators)

$\begin{array}{r}\left(\begin{array}{c}n-1\\ k-1\end{array}\right)\left(\begin{array}{c}n\\ k+1\end{array}\right)\left(\begin{array}{c}n+1\\ k\end{array}\right)=\frac{(n-1)!}{k!\left(\right(n-1)-k)!}\cdot \frac{n!}{(k-1)!(n-(k-1\left)\right)!}\cdot \frac{(n+1)!}{(k+1)!(n-k)!}\\ =\frac{(n-1)!}{k!\left(\right(n-1)-k)!}\cdot \frac{n!}{(k-1)!(n-(k-1\left)\right)!}\cdot \frac{(n+1)!}{(k+1)!\left(\right(n+1)-(k+1\left)\right)!}\\ =\left(\begin{array}{c}n-1\\ k\end{array}\right)\left(\begin{array}{c}n\\ k-1\end{array}\right)\left(\begin{array}{c}n+1\\ k+1\end{array}\right)\end{array}$