Question

4．（1）求C${\;}_{n+1}^{m}$÷（C${\;}_{n}^{m}$+C${\;}_{n}^{m-1}$）（m，n∈N^*）的值．
（2）用數(shù)學(xué)歸納法證明二項(xiàng)式定理：（a+b）ⁿ=C${\;}_{n}^{0}$aⁿ+C${\;}_{n}^{1}$a^n-1b+…+C${\;}_{n}^{r}$a^n-rb^r+…+C${\;}_{n}^{n}$bⁿ（n∈N^*，r∈N，0≤r≤n）．

Answer 1

分析 （1）利用組合數(shù)公式化簡(jiǎn)即可得出 C${\;}_{n+1}^{m}$=C${\;}_{n}^{m}$+C${\;}_{n}^{m-1}$；
（2）利用（1）的結(jié)論證明．

解答 解：（1）${C}_{n}^{m}$+${C}_{n}^{m-1}$=$\frac{n!}{m!（n-m）!}$+$\frac{n!}{（m-1）!（n-m+1）!}$=$\frac{n!（n-m+1）}{m!（n-m+1）!}$+$\frac{n!m}{m!（n-m+1）!}$
=$\frac{n!（n+1）}{m!（n-m+1）!}$=$\frac{（n+1）!}{m!（n+1-m）!}$=${C}_{n+1}^{m}$．
∴C${\;}_{n+1}^{m}$÷（C${\;}_{n}^{m}$+C${\;}_{n}^{m-1}$）=1．
（2）證明：①n=1時(shí)，左邊=a+b，右邊=${C}_{1}^{0}$a+${C}_{1}^{1}$b=a+b，
∴n=1時(shí)，等式成立．
②設(shè)n=k（k≥1，k∈N）時(shí)，（a+b）^k=${C}_{k}^{0}$a^k+C${\;}_{k}^{1}$a^k-1b+…+C${\;}_{k}^{r}$a^k-rb^r+…+C${\;}_{k}^{k}$b^k．
∴（a+b）^k+1=（${C}_{k}^{0}$a^k+C${\;}_{k}^{1}$a^k-1b+…+C${\;}_{k}^{r}$a^k-rb^r+…+C${\;}_{k}^{k}$b^k）（a+b）
=${C}_{k}^{0}$a^k+1+（${C}_{k}^{1}$+${C}_{k}^{0}$）a^kb+…+（${C}_{k}^{r}$+${C}_{k}^{r-1}$）a^k+1-rb^r+…+C${\;}_{k}^{k}$b^k+1
∵${C}_{k}^{0}$=${C}_{k+1}^{0}$，${C}_{k}^{1}$+${C}_{k}^{0}$=${C}_{k+1}^{1}$，${C}_{k}^{r}$+${C}_{k}^{r-1}$=${C}_{k+1}^{r}$，C${\;}_{k}^{k}$=${C}_{k+1}^{k+1}$，
∴（a+b）^k+1=${C}_{k+1}^{0}$a^k+1+${C}_{k+1}^{1}$a^kb+…+${C}_{k+1}^{r}$a^k+1-rb^r+…+C${\;}_{k+1}^{k+1}$b^k+1．
∴當(dāng)n=k+1時(shí)，等式成立．
綜合①②可得對(duì)任意n∈N^*，等式成立．

點(diǎn)評(píng) 本題考查了組合數(shù)公式，數(shù)學(xué)歸納法證明，屬于中檔題．