Cho k và n là hai số nguyên sao cho 4 ≤ k ≤ n. Thu gọn \(\begin{equation} \mathrm{C}_{n}^{k}+4 \mathrm{C}_{n}^{k-1}+6 \mathrm{C}_{n}^{k-2}+4 \mathrm{C}_{n}^{k-3}+\mathrm{C}_{n}^{k-4} \end{equation}\) ta được
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Lời giải:
Báo sai\(\begin{equation} \text { Từ tính chất } \mathrm{C}_{n}^{k}+\mathrm{C}_{n}^{k-1}=\mathrm{C}_{n+1}^{k} \text { với } 1 \leq k \leq n ; k, n \in \mathbb{N} . \text { Ta có } \end{equation}\)
\(\begin{equation} \begin{array}{l} \mathrm{C}_{n}^{k}+4 \mathrm{C}_{n}^{k-1}+6 \mathrm{C}_{n}^{k-2}+4 \mathrm{C}_{n}^{k-3}+\mathrm{C}_{n}^{k-4} \\ =\left(\mathrm{C}_{n}^{k}+\mathrm{C}_{n}^{k-1}\right)+3\left(\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k-2}\right)+3\left(\mathrm{C}_{n}^{k-2}+\mathrm{C}_{n}^{k-3}\right)+\left(\mathrm{C}_{n}^{k-3}+\mathrm{C}_{n}^{k-4}\right) \\ =\mathrm{C}_{n+1}^{k}+3 \mathrm{C}_{n+1}^{k-1}+3 \mathrm{C}_{n+1}^{k-2}+\mathrm{C}_{n+1}^{k-3} \\ =\left(\mathrm{C}_{n+1}^{k}+\mathrm{C}_{n+1}^{k-1}\right)+2\left(\mathrm{C}_{n+1}^{k-1}+\mathrm{C}_{n+1}^{k-2}\right)+\left(\mathrm{C}_{n+1}^{k-2}+\mathrm{C}_{n+1}^{k-3}\right) \\ =\mathrm{C}_{n+2}^{k}+2 \mathrm{C}_{n+2}^{k-1}+\mathrm{C}_{n+2}^{k-2} \\ =\left(\mathrm{C}_{n+2}^{k}+\mathrm{C}_{n+2}^{k-1}\right)+\left(\mathrm{C}_{n+2}^{k-1}+\mathrm{C}_{n+2}^{k-2}\right) \\ =\mathrm{C}_{n+3}^{k}+\mathrm{C}_{n+3}^{k-1}=\mathrm{C}_{n+4}^{k}. \end{array} \end{equation}\)
