Cho k và n là hai số tự nhiên sao cho k + 3 ≤ n. Khi đó \(\begin{equation} 2 \mathrm{C}_{n}^{k}+5 \mathrm{C}_{n}^{k+1}+4 \mathrm{C}_{n}^{k+2}+\mathrm{C}_{n}^{k+3} \end{equation}\) bằng với:
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Lời giải:
Báo sai\(\begin{equation} \text { Từ tính chất } \mathrm{C}_{n}^{k}+\mathrm{C}_{n}^{k-1}=\mathrm{C}_{n+1}^{k} \text { với } 1 \leq k \leq n ; k, n \in \mathbb{N} \text { . Ta có } \end{equation}\)
\(\begin{equation} \begin{array}{l} 2 \mathrm{C}_{n}^{k}+5 \mathrm{C}_{n}^{k+1}+4 \mathrm{C}_{n}^{k+2}+\mathrm{C}_{n}^{k+3} \\ =2\left(\mathrm{C}_{n}^{k}+\mathrm{C}_{n}^{k+1}\right)+3\left(\mathrm{C}_{n}^{k+1}+\mathrm{C}_{n}^{k+2}\right)+\left(\mathrm{C}_{n}^{k+2}+\mathrm{C}_{n}^{k+3}\right) \\ =2 \mathrm{C}_{n+1}^{k+1}+3 \mathrm{C}_{n+1}^{k+2}+\mathrm{C}_{n+1}^{k+3} \\ =2\left(\mathrm{C}_{n+1}^{k+1}+\mathrm{C}_{n+1}^{k+2}\right)+\left(\mathrm{C}_{n+1}^{k+2}+\mathrm{C}_{n+1}^{k+3}\right) \\ =2 \mathrm{C}_{n+2}^{k+2}+\mathrm{C}_{n+2}^{k+3} \\ =\mathrm{C}_{n+2}^{k+2}+\left(\mathrm{C}_{n+2}^{k+2}+\mathrm{C}_{n+2}^{k+3}\right) \\ =\mathrm{C}_{n+2}^{k+2}+\mathrm{C}_{n+3}^{k+3} \end{array} \end{equation}\)
