Rút gọn \(A=\mathrm{C}_{n}^{k}+5 \mathrm{C}_{n}^{k-1}+10 \mathrm{C}_{n}^{k-2}+10 \mathrm{C}_{n}^{k-3}+5 \mathrm{C}_{n}^{k-4}+\mathrm{C}_{n}^{k-5}\) ta được
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiTa có:
\(\begin{aligned} A &=\left(\mathrm{C}_{n}^{k}+\mathrm{C}_{n}^{k-1}\right)+4\left(\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k-2}\right)+6\left(\mathrm{C}_{n}^{k-2}+\mathrm{C}_{n}^{k-3}\right)+4\left(\mathrm{C}_{n}^{k-3}+\mathrm{C}_{n}^{k-4}\right)+\left(\mathrm{C}_{n}^{k-4}+\mathrm{C}_{n}^{k-5}\right) \\ &=\mathrm{C}_{n+1}^{k}+4 \mathrm{C}_{n+1}^{k-1}+6 \mathrm{C}_{n+1}^{k-2}+4 \mathrm{C}_{n+1}^{k-3}+\mathrm{C}_{n+1}^{k-4} \\ &=\left(\mathrm{C}_{n+1}^{k}+\mathrm{C}_{n+1}^{k-1}\right)+3\left(\mathrm{C}_{n+1}^{k-1}+\mathrm{C}_{n+1}^{k-2}\right)+3\left(\mathrm{C}_{n+1}^{k-2}+\mathrm{C}_{n+1}^{k-3}\right)+\left(\mathrm{C}_{n+1}^{k-3}+\mathrm{C}_{n+1}^{k-4}\right) \\ &=\mathrm{C}_{n+2}^{k}+3 \mathrm{C}_{n+2}^{k-1}+3 \mathrm{C}_{n+2}^{k-2}+\mathrm{C}_{n+2}^{k-3} \\ &=\left(\mathrm{C}_{n+2}^{k}+\mathrm{C}_{n+2}^{k-1}\right)+2\left(\mathrm{C}_{n+2}^{k-1}+\mathrm{C}_{n+2}^{k-2}\right)+\left(\mathrm{C}_{n+2}^{k-2}+\mathrm{C}_{n+2}^{k-3}\right) \\ &=\mathrm{C}_{n+3}^{k}+2 \mathrm{C}_{n+3}^{k-1}+\mathrm{C}_{n+3}^{k-2} \\ &=\left(\mathrm{C}_{n+3}^{k}+\mathrm{C}_{n+3}^{k-1}\right)+\left(\mathrm{C}_{n+3}^{k-1}+\mathrm{C}_{n+3}^{k-2}\right) \\ &=\mathrm{C}_{n+4}^{k}+\mathrm{C}_{n+4}^{k-1}=\mathrm{C}_{n+5}^{k} \end{aligned}\)
