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和式

记号

符号:\(\huge\sum\)

eg.

1. \(a_1 + a_2 + \cdots + a_{k-1} + a_k + a_{k+1}+\cdots +a_{n-1}+a_n = \sum_{k=1}^na_k=\sum_{1\leq k \leq n} a_k\)
2. \(\sum_{\substack{1\leq k\leq n \\ \text{k } prime}}\)

成套方法

解决将和式转为封闭式的方法

前\(n\)个自然数的和

命题

将\(\sum_{k=1}^nk转为封闭式\)

求解

方法:成套方法

1. 转为递归式

令\(S(n)=\sum_{k=1}^nk\)
不难看出,\(S(n)=S(n-1)+n\)

1. 一般化

令\(R(n)\)为\(S(n)\)的一般形式
即\(R(0)=\alpha \qquad R(n)=R(n-1)+\beta n+\gamma\)

(1) 令\(R(n)=1\)

\[\therefore R(0)=1\]

\[\therefore \alpha = 1\]

\[\because R(n)=R(n-1)+\beta n+\gamma\]

\[\therefore 1=1+\beta n + \gamma\]

\[ \left\{ \begin{aligned} \alpha = 1 \\beta = 0 \\gamma = 0 \end{aligned} \right. \]

(2) 令\(R(n)=n\)

\[\therefore R(0) = 0\]

\[\therefore \alpha = 0\]

\[\because R(n)=R(n-1)+\beta n+\gamma\]

\[\therefore n = (n-1)+\beta n + \gamma\]

\[ \left\{ \begin{aligned} \alpha = 0 \\beta = 0 \\gamma = 1 \end{aligned} \right. \]

(3) 令\(R(n) = n^2\)

\[\therefore R(0) = 0\]

\[\therefore \alpha = 0\]

\[\because R(n)=R(n-1)+\beta n+\gamma\]

\[\therefore n^2 = (n-1)^2+\beta n + \gamma\]

\[\therefore n^2 = n^2 - 2n + 1+\beta n + \gamma\]

\[\therefore -1 =(\beta - 2) n + \gamma\]

\[ \left\{ \begin{aligned} \alpha = 0 \\beta = 2 \\gamma = -1 \end{aligned} \right. \]

3.计算系数

令\(R(n)=x\alpha + y\beta + z\theta\)

(1) 当\(R(n) = 1\)时:

\[\because\left\{ \begin{aligned} \alpha = 1 \\beta = 0 \\gamma = 0 \end{aligned} \right. \]

\[\therefore x = 1\]

(2) 当\(R(n) = n\)时:

\[\because\left\{ \begin{aligned} \alpha = 0 \\beta = 0 \\gamma = 1 \end{aligned} \right. \]

\[\therefore z = n\]

(3) 当\(R(n) = n^2\)时:

\[ \left\{ \begin{aligned} \alpha = 0 \\beta = 2 \\gamma = -1 \end{aligned} \right. \]

\[\therefore 2y - z = n^2\]

综上:

\[ \left\{ \begin{aligned} x = 1 \z = n \2y - z = n^2 \end{aligned} \right. \]

解得
\[ \left\{ \begin{aligned} x = 1 \y = \frac{n\cdot (n+1)}{2} \z = n \end{aligned} \right. \]

4.具体化

\[S(n) = S(n-1) + n\]

令\(P(n)\)为当\(\beta = 1, \gamma = 0\)时\(R(n)\)的值

\[\therefore P(n) = P(n-1) + n = S(n)\]

\(\therefore S(n)\)为当\(\beta = 1, \gamma = 0\)时\(R(n)\)的值

\[\therefore S(n) = y\]

\[\therefore S(n) = \frac{n \cdot (n+1)}{2}\]

[ 具体数学 ] 和式与封闭式

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原文地址：https://www.cnblogs.com/zhangtianli/p/12233360.html