二项式定理

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12.二项式定理2023-08-22

二项式定理#

$(a + b)^n = \sum^n_{k = 0}C^k_na^kb^{n - k}$
其中，等号右侧叫做等号左侧的二次展开式。

证明：
运用数学归纳法。

当 $n = 1$ 时，显然有

a + b = C_{1}^{0} a^{0} b^{1} + C_{1}^{1} a^{1} b^{0} = a + b

$a + b = C^0_1a^0b^1 + C^1_1a^1b^0 = a + b$

假设 $n = m$ 时，命题成立。

当 $n = m + 1$ 时，

\begin{aligned} (a + b)^{n} \\ = & (a + b)^{m + 1} \\ = & (a + b) \sum_{k = 0}^{m} C_{m}^{k} a^{k} b^{m - k} \\ = & \sum_{k = 0}^{m} C_{m}^{k} a^{k + 1} b^{m - k} + \sum_{k = 0}^{m} C_{m}^{k} a^{k} b^{m - k + 1} \\ = & \sum_{k = 1}^{m + 1} C_{m}^{k - 1} a^{k} b^{m - k + 1} + \sum_{k = 0}^{m} C_{m}^{k} a^{k} b^{m - k + 1} \\ = & \sum_{k = 0}^{m + 1} (C_{m}^{k - 1} + C_{m}^{k}) a^{k} b^{m - k + 1} \\ = & \sum_{k = 0}^{m + 1} C_{m + 1}^{k} a^{k} b^{m - k + 1} \\ = & \sum_{k = 0}^{m + 1} C_{m + 1}^{k} a^{k} b^{m - k + 1} \\ = & \sum_{k = 0}^{n} C_{n}^{k} a^{k} b^{n - k} \end{aligned}

$\begin{aligned} & (a + b)^n \\ = & (a + b)^{m + 1} \\ = & (a + b)\sum^m_{k = 0}C_m^ka^kb^{m - k} \\ = & \sum^m_{k = 0}C^k_ma^{k + 1}b^{m - k} + \sum^m_{k = 0}C_m^ka^kb^{m - k + 1} \\ = & \sum^{m + 1}_{k = 1}C^{k - 1}_ma^kb^{m - k + 1} + \sum^m_{k = 0}C_m^ka^kb^{m - k + 1} \\ = & \sum^{m + 1}_{k = 0}(C^{k - 1}_m + C^k_m)a^kb^{m - k + 1} \\ = & \sum^{m + 1}_{k = 0}C_{m + 1}^{k}a^kb^{m - k + 1} \\ = & \sum^{m + 1}_{k = 0}C^k_{m + 1}a^kb^{m - k + 1} \\ = & \sum^n_{k = 0}C^k_na^kb^{n - k} \end{aligned}$

证毕。

作者：白简

出处：https://www.cnblogs.com/baijian0212/articles/erxiangshi.html

版权：本作品采用「署名-非商业性使用-相同方式共享 4.0 国际」许可协议进行许可。