Tìm giá trị n Î ℕ thỏa mãn: \(C_{n + 1}^1 + 3C_{n + 2}^2 = C_{n + 1}^3\).

\(C_{n + 1}^1 + 3C_{n + 2}^2 = C_{n + 1}^3\)

\( \Leftrightarrow n + 1 + 3\,.\,\frac{{\left( {n + 2} \right)!}}{{2!\,.\,n!}} = \frac{{\left( {n + 1} \right)!}}{{3!\,.\,\left( {n - 2} \right)!}}\)

\( \Leftrightarrow n + 1 + \frac{3}{2}\,.\,\left( {n + 2} \right)\left( {n + 1} \right) = \frac{1}{6}\,.\,\left( {n + 1} \right)n\left( {n - 1} \right)\)

\( \Leftrightarrow n + 1 + \frac{3}{2}\,.\,\left( {{n^2} + 3n + 2} \right) = \frac{1}{6}\,.\,\left( {{n^3} - n} \right)\)

\( \Leftrightarrow n + 1 + \frac{3}{2}{n^2} + \frac{9}{2}n + 3 = \frac{1}{6}{n^3} - \frac{1}{6}n\)

\( \Leftrightarrow \frac{1}{6}{n^3} - \frac{3}{2}{n^2} - \frac{{17}}{3}n - 4 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}n = 12\;\left( n \right)\\n = - 1\;\left( l \right)\\n = - 2\;\left( l \right)\end{array} \right.\)

Vậy n = 12 là số nguyên dương cần tìm.