Cho \(C_{x + 1}^y:C_x^{y + 1}:C_x^{y - 1} = 6:5:2\). Khi đó tổng $x + y$ bằng:

ĐK: \(\left\{ \begin{array}{l}x + 1 \ge y \ge 0\\x \ge y + 1 \ge 0\\x \ge y - 1 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y \ge 1\\x \ge y + 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}y \ge 1\\x \ge 2\end{array} \right.\,\,\left( {x,y \in N} \right)\)

\(\begin{array}{l}C_{x + 1}^y:C_x^{y + 1}:C_x^{y - 1} = 6:5:2 \Rightarrow \left\{ \begin{array}{l}\dfrac{{C_{x + 1}^y}}{{C_x^{y + 1}}} = \dfrac{6}{5}\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\\dfrac{{C_{x + 1}^y}}{{C_x^{y - 1}}} = \dfrac{6}{2} = 3\,\,\left( 2 \right)\end{array} \right.\\\left( 1 \right) \Leftrightarrow \dfrac{{\dfrac{{\left( {x + 1} \right)!}}{{y!\left( {x + 1 - y} \right)!}}}}{{\dfrac{{x!}}{{\left( {y + 1} \right)!\left( {x - y - 1} \right)!}}}} = \dfrac{6}{5}\\ \Leftrightarrow \dfrac{{\left( {x + 1} \right)!}}{{y!\left( {x + 1 - y} \right)!}}.\dfrac{{\left( {y + 1} \right)!\left( {x - y - 1} \right)!}}{{x!}} = \dfrac{6}{5}\\ \Leftrightarrow \dfrac{{\left( {x + 1} \right)\left( {y + 1} \right)}}{{\left( {x - y} \right)\left( {x - y + 1} \right)}} = \dfrac{6}{5}\,\,\,\left( 3 \right)\\\left( 2 \right) \Leftrightarrow \dfrac{{\dfrac{{\left( {x + 1} \right)!}}{{y!\left( {x + 1 - y} \right)!}}}}{{\dfrac{{x!}}{{\left( {y - 1} \right)!\left( {x - y + 1} \right)!}}}} = 3\\ \Leftrightarrow \dfrac{{\left( {x + 1} \right)!}}{{y!\left( {x + 1 - y} \right)!}}\dfrac{{\left( {y - 1} \right)!\left( {x - y + 1} \right)!}}{{x!}} = 3\\ \Leftrightarrow \dfrac{{x + 1}}{y} = 3 \Rightarrow x = 3y - 1\end{array}\)

Thay vào (3) ta có:

\(\begin{array}{l}\dfrac{{3y\left( {y + 1} \right)}}{{\left( {2y - 1} \right)2y}} = \dfrac{6}{5}\\ \Leftrightarrow \dfrac{{y + 1}}{{4y - 2}} = \dfrac{2}{5} \Leftrightarrow 5y + 5 = 8y - 4\\ \Leftrightarrow 3y = 9 \Leftrightarrow y = 3\,\,\left( {tm} \right) \Rightarrow x = 8\,\,\left( {tm} \right)\\ \Rightarrow x + y = 11\end{array}\)