[ 具体数学 ] 3:和式与封闭式

成套方法

解决将和式转为封闭式的方法

命题

将${\sum }_{k=1}^{n}k$转为封闭式

求解

方法:成套方法

1. 转为递归式

令$S(n)={\sum }_{k=1}^{n}k$
不难看出,$S(n)=S(n-1)+n$

2. 一般化

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

(1) 令$R(n)=1$

$$\begin{gathered} \therefore R(0)=1 \\ \therefore \alpha =1 \\ \because R(n)=R(n-1)+\beta n+\gamma \\ \therefore 1=1+\beta n+\gamma \end{gathered}$$

$$\{\begin{array}{r}\alpha =1\\ \beta =0\\ \gamma =0\end{array}$$

(2) 令$R(n)=n$

$$\begin{gathered} \therefore R(0)=0 \\ \therefore \alpha =0 \\ \because R(n)=R(n-1)+\beta n+\gamma \\ \therefore n=(n-1)+\beta n+\gamma \end{gathered}$$

$$\{\begin{array}{r}\alpha =0\\ \beta =0\\ \gamma =1\end{array}$$

(3) 令$R(n)=n^2$

$$\begin{gathered} \therefore R(0)=0 \\ -\therefore \alpha =0 \\ \because R(n)=R(n-1)+\beta n+\gamma \\ \therefore n^2=(n-1{)}^2+\beta n+\gamma \\ \therefore n^2=n^2-2n+1+\beta n+\gamma \\ \therefore -1=(\beta -2)n+\gamma \end{gathered}$$

$$\{\begin{array}{r}\alpha =0\\ \beta =2\\ \gamma =-1\end{array}$$

3.计算系数

令$R(n)=A(n)\alpha +B(n)\beta +C(n)\gamma$

(1) 当$R(n)=1$时:

$$\begin{gathered} \because \{\begin{array}{r}\alpha =1\\ \beta =0\\ \gamma =0\end{array} \\ \therefore A(n)=1 \end{gathered}$$

(2) 当$R(n)=n$时:

$$\begin{gathered} \because \{\begin{array}{r}\alpha =0\\ \beta =0\\ \gamma =1\end{array} \\ \therefore C(n)=n \end{gathered}$$

(3) 当$R(n)=n^2$时:

$$\begin{gathered} \{\begin{array}{r}\alpha =0\\ \beta =2\\ \gamma =-1\end{array} \\ \therefore 2B(n)-C(n)=n^2 \end{gathered}$$

综上:

$$\{\begin{array}{r}A(n)=1\\ C(n)=n\\ 2B(n)-C(n)=n^2\end{array}$$

解得
$$\{\begin{array}{r}A(n)=1\\ B(n)=\frac{n\cdot (n+1)}{2}\\ C(n)=n\end{array}$$

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

$$\begin{gathered} \therefore S(n)=B(n) \\ \therefore S(n)=\frac{n\cdot (n+1)}{2} \end{gathered}$$