Cho các số nguyên dương p ≤ n. Thu gọn \(\mathrm{C}_{n}^{1}+2 \cdot \frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{n}^{1}}+3 \cdot \frac{\mathrm{C}_{n}^{3}}{\mathrm{C}_{n}^{2}}+\cdots+p \cdot \frac{\mathrm{C}_{n}^{p}}{\mathrm{C}_{n}^{p-1}}+\cdots+n \cdot \frac{\mathrm{C}_{n}^{n}}{\mathrm{C}_{n}^{n-1}}\) ta được
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Lời giải:
Báo saiTa có:
\(\begin{aligned} \mathrm{C}_{n}^{1} &=n ; \\ 2 \cdot \frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{n}^{1}} &=2 \cdot \frac{\frac{n !}{2 !(n-2) !}}{n}=\frac{n !}{(n-2) ! n}=n-1 ; \end{aligned}\)
\(3 \cdot \frac{\mathrm{C}_{n}^{3}}{\mathrm{C}_{n}^{2}}=3 \cdot \frac{\frac{n !}{3 !(n-3) !}}{\frac{n !}{2 !(n-2) !}}=\frac{(n-2) !}{(n-3) !}=n-2\)
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\(p \cdot \frac{\mathrm{C}_{n}^{p}}{\mathrm{C}_{n}^{p-1}}=p \cdot \frac{\frac{n !}{p !(n-p) !}}{\frac{n !}{(p-1) !(n-p+1) !}}=p \cdot \frac{(p-1) !(n-p+1) !}{p !(n-p) !}=p \cdot \frac{n-p+1}{p}=n-p+1 ;\)
\(n \cdot \frac{\mathrm{C}_{n}^{n}}{\mathrm{C}_{n}^{n-1}}=n \cdot \frac{1}{\frac{n !}{1 !(n-1) !}}=1\)
Do đó \(\mathrm{C}_{n}^{1}+2 \cdot \frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{n}^{1}}+3 \cdot \frac{\mathrm{C}_{n}^{3}}{\mathrm{C}_{n}^{2}}+\cdots+p \cdot \frac{\mathrm{C}_{n}^{p}}{\mathrm{C}_{n}^{p-1}}+\cdots+n \cdot \frac{\mathrm{C}_{n}^{n}}{\mathrm{C}_{n}^{n-1}}=n+(n-1)+(n-2)+\cdots+(n-p+1)+\cdots+1=\frac{n(n+1)}{2}\)
