Rút gọn \(A=\mathrm{C}_{n}^{1}+2 \cdot \frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{n}^{1}}+3 \cdot \frac{\mathrm{C}_{n}^{3}}{\mathrm{C}_{n}^{2}}+\cdots+n \cdot \frac{\mathrm{C}_{n}^{n}}{\mathrm{C}_{n}^{n-1}}\)
Suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo sai\(\begin{array}{l} \mathrm{C}_{n}^{1}=n \\ 2 \cdot \frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{n}^{1}}=2 \cdot \frac{\frac{2 !(n-2) !}{n !}}{\frac{(n-1) !}{n !}}=2 \cdot \frac{n !}{2 !(n-2) !} \cdot \frac{(n-1) !}{n !}=n-1 . \\ \frac{\mathrm{C}_{n}^{3}}{\mathrm{C}_{n}^{2}}=3 \cdot \frac{\frac{3 !(n-3) !}{n !}}{\frac{2 !}{2 !(n-2) !}}=n-2 \end{array}\)
....
\(k \cdot \frac{\mathrm{C}_{n}^{k}}{\mathrm{C}_{n}^{k-1}}=k \cdot \frac{\frac{n !}{k !(n-k) !}}{\frac{n !}{(k-1) !(n-k+1) !}}=n-k+1\)
...
\(n \cdot \frac{\mathrm{C}_{n}^{n}}{\mathrm{C}_{n}^{n-1}}=n \cdot \frac{1}{\frac{n !}{(n-1) ! 1 !}}=1\)
cộng vế theo vế ta được
\(\begin{aligned} A &=n+(n-1)+(n-2)+\cdots+(n-k+1)+\cdots+1 \\ &=1+2+3+\cdots+n=\frac{n(n+1)}{2} \end{aligned}\)
