Question

Let\(n\)be a positive integer. Show that\(\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 2}\\{n + 1}\end{array}} \right)/2\).

Short answer

The expression\(\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 2}\\{n + 1}\end{array}} \right)/2\) is proved by the Pascal identity.

Step-by-step solution

Formula of Pascal identity

Pascal identity:

\(\left( {\begin{array}{*{20}{c}}{n + 1}\\k\end{array}} \right) = \left( {\begin{array}{*{20}{c}}n\\{k - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}n\\k\end{array}} \right)\)

Use the Pascal identity to prove the expression

Let \({\bf{n}}\) be a positive integer.

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 1}\\{n + 1}\end{array}} \right)\end{array}\)

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 1)!}}{{(n + 1)!(2n + 1 - (n + 1))!}}\\\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 1)!}}{{(n + 1)!n!}}\end{array}\)

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 1)!}}{{(n + 1)!n!}} \cdot \frac{{2n + 2}}{{2n + 2}}\\\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 2)!}}{{(n + 1)!n!}} \cdot \frac{1}{{2n + 2}}\end{array}\)

\(\begin{array}{l}\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 2)!}}{{(n + 1)!n!}} \cdot \frac{1}{{2(n + 1)}}\\\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \frac{{(2n + 2)!}}{{(n + 1)!(n + 1)!}} \cdot \frac{1}{2}\end{array}\)

\(\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 2}\\{n + 1}\end{array}} \right) \cdot \frac{1}{2}\)

\(\left( {\begin{array}{*{20}{c}}{2n}\\{n + 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{2n}\\n\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{2n + 2}\\{n + 1}\end{array}} \right)/2\)