和式
记号
符号:\(\huge\sum\)
eg.
- \(a_1 + a_2 + \cdots + a_{k-1} + a_k + a_{k+1}+\cdots +a_{n-1}+a_n = \sum_{k=1}^na_k=\sum_{1\leq k \leq n} a_k\)
- \(\sum_{\substack{1\leq k\leq n \\ \text{k } prime}}\)
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
- 转为递归式
令\(S(n)=\sum_{k=1}^nk\)
不难看出,\(S(n)=S(n-1)+n\)
- 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
\[\therefore S(n) = y\]
\[\therefore S(n) = \frac{n \cdot (n+1)}{2}\]