[ 具体数学 ] 2:和式与封闭式

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

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)=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}

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