[ 具体数学 ] 和式与封闭式
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和式
记号
符号:(hugesum)
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_{1leq k leq n} a_k)
- (sum_{substack{1leq kleq n \ ext{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)+eta n+gamma)
(1) 令(R(n)=1)
[ herefore R(0)=1]
[ herefore alpha = 1]
[ecause R(n)=R(n-1)+eta n+gamma]
[ herefore 1=1+eta n + gamma]
[ left{ egin{aligned} alpha = 1 \beta = 0 \gamma = 0 end{aligned} ight. ]
(2) 令(R(n)=n)
[ herefore R(0) = 0]
[ herefore alpha = 0]
[ecause R(n)=R(n-1)+eta n+gamma]
[ herefore n = (n-1)+eta n + gamma]
[ left{ egin{aligned} alpha = 0 \beta = 0 \gamma = 1 end{aligned} ight. ]
(3) 令(R(n) = n^2)
[ herefore R(0) = 0]
[ herefore alpha = 0]
[ecause R(n)=R(n-1)+eta n+gamma]
[ herefore n^2 = (n-1)^2+eta n + gamma]
[ herefore n^2 = n^2 - 2n + 1+eta n + gamma]
[ herefore -1 =(eta - 2) n + gamma]
[ left{ egin{aligned} alpha = 0 \beta = 2 \gamma = -1 end{aligned} ight. ]
3.计算系数
令(R(n)=xalpha + yeta + z heta)
(1) 当(R(n) = 1)时:
[ecauseleft{ egin{aligned} alpha = 1 \beta = 0 \gamma = 0 end{aligned} ight. ]
[ herefore x = 1]
(2) 当(R(n) = n)时:
[ecauseleft{ egin{aligned} alpha = 0 \beta = 0 \gamma = 1 end{aligned} ight. ]
[ herefore z = n]
(3) 当(R(n) = n^2)时:
[ left{ egin{aligned} alpha = 0 \beta = 2 \gamma = -1 end{aligned} ight. ]
[ herefore 2y - z = n^2]
综上:
[ left{ egin{aligned} x = 1 z = n 2y - z = n^2 end{aligned} ight. ]
解得
[ left{
egin{aligned}
x = 1 y = frac{ncdot (n+1)}{2} z = n
end{aligned}
ight.
]
4.具体化
[S(n) = S(n-1) + n]
令(P(n))为当(eta = 1, gamma = 0)时(R(n))的值
[ herefore P(n) = P(n-1) + n = S(n)]
( herefore S(n))为当(eta = 1, gamma = 0)时(R(n))的值
[ herefore S(n) = y]
[ herefore S(n) = frac{n cdot (n+1)}{2}]
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