成套方法
解决将和式转为封闭式的方法
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\(R(n)=A(n)\alpha + B(n)\beta + C(n)\gamma\)
(1) 当\(R(n) = 1\)时:
\[\because\left\{
\begin{aligned}
\alpha = 1 \\
\beta = 0 \\
\gamma = 0
\end{aligned}
\right.
\]
\[\therefore A(n) = 1
\]
(2) 当\(R(n) = n\)时:
\[\because\left\{
\begin{aligned}
\alpha = 0 \\
\beta = 0 \\
\gamma = 1
\end{aligned}
\right.
\]
\[\therefore C(n) = n
\]
(3) 当\(R(n) = n^2\)时:
\[ \left\{
\begin{aligned}
\alpha = 0 \\
\beta = 2 \\
\gamma = -1
\end{aligned}
\right.
\]
\[\therefore 2B(n) - C(n) = n^2
\]
综上:
\[ \left\{
\begin{aligned}
A(n) = 1 \\
C(n) = n \\
2B(n) - C(n) = n^2
\end{aligned}
\right.
\]
解得
\[ \left\{
\begin{aligned}
A(n) = 1 \\
B(n) = \frac{n\cdot (n+1)}{2} \\
C(n) = n
\end{aligned}
\right.
\]
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
\[\therefore S(n) = B(n)
\]
\[\therefore S(n) = \frac{n \cdot (n+1)}{2}
\]