Tìm số nguyên dương n sao cho \(C_n^1 + C_n^2 + C_n^3 = \dfrac{{7n}}{2}\)

* Hướng dẫn giải

Ta có

\(\begin{array}{l}C_n^1 + C_n^2 + C_n^3 = \dfrac{{7n}}{2}\\ \Leftrightarrow \dfrac{{n!}}{{\left( {n - 1} \right)!}} + \dfrac{{n!}}{{2!.\left( {n - 2} \right)!}} + \dfrac{{n!}}{{3!.\left( {n - 3} \right)!}} = \dfrac{{7n}}{2}\\ \Leftrightarrow n + \dfrac{{n\left( {n - 1} \right)}}{2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} = \dfrac{{7n}}{2}\\ \Leftrightarrow 2 + n - 1 + \dfrac{{{n^2} - 3n + 2}}{3} = 7\\ \Leftrightarrow {n^2} = 16\\ \Leftrightarrow n = 4(n > 0)\end{array}\)