INFO

若无特殊说明，本章涉及的变量皆为正整数。

简介

$$(a+b{)}^n=\sum _{i=0}^n(\genfrac{}{}{0ex}{}{n}{i})a^{n-i}{b}^i$$

证明

使用数学归纳法。

设 $n=\phantom{0}$$\phantom{0}k$ 时二项式定理成立，考察 $n=\phantom{0}$$\phantom{0}k+\phantom{0}$$\phantom{0}1$ 时是否也成立：

$$\begin{array}{rl}& =(a+b{)}^{k+1}\\ & =(a+b)\cdot (a+b{)}^k\\ & =a(a+b{)}^k+b(a+b{)}^k\\ & =a\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})a^{k-i}{b}^i+b\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})a^{k-j}{b}^j\\ & =\sum _{i=0}^k(\genfrac{}{}{0ex}{}{k}{i})a^{k-i+1}{b}^i+\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})a^{k-j}{b}^{j+1}& & \text{将 }a,b\text{ 乘进去}\\ & =a^{k+1}+\sum _{i=1}^k(\genfrac{}{}{0ex}{}{k}{i})a^{k-i+1}{b}^i+\sum _{j=0}^k(\genfrac{}{}{0ex}{}{k}{j})a^{k-j}{b}^{j+1}& & \text{提出 }i=0\text{ 的项}\\ & =a^{k+1}+\sum _{i=1}^k(\genfrac{}{}{0ex}{}{k}{i})a^{k-i+1}{b}^i+\sum _{\lambda =1}^{k+1}(\genfrac{}{}{0ex}{}{k}{\lambda -1})a^{k-\lambda +1}{b}^{\lambda }& & \text{设 }\lambda =j+1\text{，代入}\\ & =a^{k+1}+\sum _{i=1}^k(\genfrac{}{}{0ex}{}{k}{i})a^{k-i+1}{b}^i+b^{k+1}+\sum _{\lambda =1}^k(\genfrac{}{}{0ex}{}{k}{\lambda -1})a^{k-\lambda +1}{b}^{\lambda }& & \text{提出 }\lambda =k+1\text{ 的项}\\ & =a^{k+1}+b^{k+1}+\sum _{i=1}^k(\genfrac{}{}{0ex}{}{k+1}{i})a^{k+1-i}{b}^i& & \text{套用帕斯卡法则}\\ & =\sum _{i=0}^{k+1}(\genfrac{}{}{0ex}{}{k+1}{i})a^{k+1-i}{b}^i\end{array}$$

$\therefore$ 二项式定理满足递推成立关系：

$$n=k\text{ 时成立}⟹n=k+1\text{ 时成立}$$

$\because n=\phantom{0}$$\phantom{0}1$ 时 $(a+\phantom{0}$$\phantom{0}b{)}^1=\phantom{0}$$\phantom{0}\sum _{i=0}^1(\genfrac{}{}{0ex}{}{n}{i})a^{n-i}{b}^i=\phantom{0}$$\phantom{0}a+\phantom{0}$$\phantom{0}b$ 成立，

$\therefore$ 二项式定理在 $n=\phantom{0}$$\phantom{0}1$ 之后的任何整数都成立。