Question

1．解不等式組$\left\{\begin{array}{l}{{{3}^{k}C}_{5}^{k}≥{3}^{k-1}{C}_{5}^{k-1}}\\{{{3}^{k}C}_{5}^{k}{{≥3}^{k+1}C}_{5}^{k+1}}\end{array}\right.$．

Answer 1

分析 直接利用組合數(shù)的性質(zhì)化簡(jiǎn)不等式組，求解即可．

解答 解：不等式組$\left\{\begin{array}{l}{{{3}^{k}C}_{5}^{k}≥{3}^{k-1}{C}_{5}^{k-1}}\\{{{3}^{k}C}_{5}^{k}{{≥3}^{k+1}C}_{5}^{k+1}}\end{array}\right.$可得：$\left\{\begin{array}{l}{3C}_{5}^{k}≥{C}_{5}^{k-1}\\{C}_{5}^{k}{≥3C}_{5}^{k+1}\end{array}\right.$，即：$\left\{\begin{array}{l}\frac{3×51}{k!（5-k）！}≥\frac{51}{（k-1）!（6-k）！}\\ \frac{51}{k!（5-k）！}≥\frac{3×51}{（k+1）!（4-k）！}\end{array}\right.$
可得：$\left\{\begin{array}{l}\frac{3}{k}≥\frac{1}{6-k}\\ \frac{1}{5-k}≥\frac{3}{k+1}\end{array}\right.$
解得：$\frac{7}{2}≤k≤\frac{9}{2}$，k∈N，
可得k=4．

點(diǎn)評(píng) 本題考查組合數(shù)公式的應(yīng)用，不等式組的解法，考查計(jì)算能力．