And style
mark
Symbols: \ (\ 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}} \)
Complete Method
Solve the method and style into closed
Before \ (n \) natural numbers and
Proposition
The \ (\ sum_ {k = 1 } ^ nk Switch Closed \)
Solving
Method: Complete Method
- Into recursive
Order \ (S (n) = \
sum_ {k = 1} ^ nk \) easy to see, \ (S (n-) = S (n--. 1) n-+ \)
- generalization
Order \ (R (n) \) of \ (S (n) \) in the general form
i.e. \ (R (0) = \ alpha \ qquad R (n) = R (n-1) + \ beta n + \ gamma \)
(1) Order \ (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) Order \ (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) Order \ (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. Calculation Coefficient
令\(R(n)=x\alpha + y\beta + z\theta\)
(1) When \ (R (n) = 1 \) when:
\[\because\left\{ \begin{aligned} \alpha = 1 \\ \beta = 0 \\ \gamma = 0 \end{aligned} \right. \]
\[\therefore x = 1\]
(2) When \ (R (n) = n \) when:
\[\because\left\{ \begin{aligned} \alpha = 0 \\ \beta = 0 \\ \gamma = 1 \end{aligned} \right. \]
\[\therefore z = n\]
(3) When \ (R (n) = n ^ 2 \) when:
\[ \left\{ \begin{aligned} \alpha = 0 \\ \beta = 2 \\ \gamma = -1 \end{aligned} \right. \]
\[\therefore 2y - z = n^2\]
In summary:
\[ \left\{ \begin{aligned} x = 1 \\ z = n \\ 2y - z = n^2 \end{aligned} \right. \]
解得
\[ \left\{ \begin{aligned} x = 1 \\ y = \frac{n\cdot (n+1)}{2} \\ z = n \end{aligned} \right. \]
4. embodying
\[S(n) = S(n-1) + n\]
Order \ (P (n) \) that when \ (\ beta = 1, \ gamma = 0 \) when \ (R (n) \) values
\[\therefore P(n) = P(n-1) + n = S(n)\]
\ (\ therefore S (n) \) that when \ (\ beta = 1, \ gamma = 0 \) when (R (n) \) \ value
\[\therefore S(n) = y\]
\[\therefore S(n) = \frac{n \cdot (n+1)}{2}\]