閱讀指南
編輯
基礎知識
編輯
數列的定義與基本概念
編輯
按照一定的順序排列的一列數叫做數列 （Sequence of number ），數列中的每個數叫做這個數列的一個項 （term ），數列中的每一個數都和它的序號有關，排在第1位的數稱為這個數列的第1項（也叫做首項），排在第2位的數叫做數列的第2項……以此類推，排在第n位的數叫做這個數列的第n項 。所以，數列的一般形式可以寫成：
a
1
,
a
2
,
a
3
,
⋯
,
a
n
,
⋯
{\displaystyle a_{1},a_{2},a_{3},\cdots ,a_{n},\cdots }
{
a
n
}
{\displaystyle \left\{a_{n}\right\}}
項目有限的數列叫做有窮數列 ，而項目無窮的數列則稱之為無窮數列 。項目全部為正數的數列叫做正項數列 。
如果數列
{
a
n
}
{\displaystyle \left\{a_{n}\right\}}
一般項公式 （general term ）或通項公式 。我們可以通過這個公式計算出數列的各個項。
我們將
a
1
+
a
2
+
a
3
+
⋯
+
a
n
{\displaystyle a_{1}+a_{2}+a_{3}+\cdots +a_{n}}
{
a
n
}
{\displaystyle \{a_{n}\}}
S
n
{\displaystyle S_{n}}
S
n
=
a
1
+
a
2
+
⋯
+
a
n
{\displaystyle S_{n}=a_{1}+a_{2}+\cdots +a_{n}}
{
n
(
n
+
1
)
10
}
{\displaystyle \{{\frac {n(n+1)}{10}}\}}
{
a
n
}
{\displaystyle \{a_{n}\}}
a
n
=
n
2
−
n
+
1
{\displaystyle a_{n}=n^{2}-n+1}
a
n
=
n
(
n
−
1
)
2
{\displaystyle a_{n}={\frac {n(n-1)}{2}}}
a
n
=
n
(
n
+
1
)
2
{\displaystyle a_{n}={\frac {n(n+1)}{2}}}
a
n
=
n
(
n
+
2
)
2
{\displaystyle a_{n}={\frac {n(n+2)}{2}}}
a
n
=
1
n
+
1
n
+
1
+
1
n
+
2
+
1
n
+
3
+
.
.
.
+
1
n
2
(
n
∈
N
+
)
{\displaystyle a_{n}={\frac {1}{n}}+{\frac {1}{n+1}}+{\frac {1}{n+2}}+{\frac {1}{n+3}}+...+{\frac {1}{n^{2}}}\quad (n\in \mathbb {N} ^{+})}
a
2
=
{\displaystyle a_{2}=}
1
2
{\displaystyle {\frac {1}{2}}}
1
2
+
1
3
{\displaystyle {\frac {1}{2}}+{\frac {1}{3}}}
1
2
+
1
3
+
1
4
{\displaystyle {\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}}
1
2
+
1
3
+
1
4
+
1
5
{\displaystyle {\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}}
a
n
=
1
n
+
1
+
1
n
+
2
+
1
n
+
3
+
1
n
+
4
+
.
.
.
+
1
2
n
(
n
∈
N
+
)
{\displaystyle a_{n}={\frac {1}{n+1}}+{\frac {1}{n+2}}+{\frac {1}{n+3}}+{\frac {1}{n+4}}+...+{\frac {1}{2n}}\quad (n\in \mathbb {N} ^{+})}
a
n
+
1
−
a
n
=
{\displaystyle a_{n+1}-a_{n}=}
1
2
n
+
1
{\displaystyle {\frac {1}{2n+1}}}
1
2
n
+
2
{\displaystyle {\frac {1}{2n+2}}}
1
2
n
+
1
+
1
2
n
+
2
{\displaystyle {\frac {1}{2n+1}}+{\frac {1}{2n+2}}}
1
2
n
+
1
+
1
2
n
+
2
{\displaystyle {\frac {1}{2n+1}}+{\frac {1}{2n+2}}}
常見性質與分類
編輯
{
a
n
}
{\displaystyle \{a_{n}\}}
a
n
=
n
3
n
+
1
{\displaystyle a_{n}={\frac {n}{3n+1}}}
遞推關係式及其簡單使用
編輯
如果已知一個數列
{
a
n
}
{\displaystyle \{a_{n}\}}
遞推關係式 （recurrence relation ），簡稱遞推式 。以遞推式描述的數列也叫做遞推數列 ，它的每一項需要由給定的前幾項以及遞推關係式共同確定。
{
a
n
}
{\displaystyle \{a_{n}\}}
a
1
=
1
,
a
n
+
1
=
1
2
a
n
+
1
2
n
{\displaystyle a_{1}=1,a_{n+1}={\frac {1}{2}}a_{n}+{\frac {1}{2n}}}
{
a
n
}
{\displaystyle \{a_{n}\}}
a
1
=
2
,
a
2
=
5
,
a
n
+
1
=
a
n
+
2
+
a
n
(
n
≥
3
,
n
∈
N
)
{\displaystyle a_{1}=2,a_{2}=5,a_{n+1}=a_{n+2}+a_{n}\quad (n\geq 3,n\in \mathbb {N} )}
{
a
n
}
{\displaystyle \{a_{n}\}}
a
1
=
1
,
a
n
+
a
n
+
1
=
2
(
n
∈
N
+
)
{\displaystyle a_{1}=1,a_{n}+a_{n+1}=2\quad (n\in \mathbb {N} ^{+})}
{
a
n
}
{\displaystyle \{a_{n}\}}
a
1
=
1
,
a
n
=
n
(
a
n
+
1
−
a
n
)
(
n
∈
N
+
)
{\displaystyle a_{1}=1,a_{n}=n(a_{n+1}-a_{n})\quad (n\in \mathbb {N} ^{+})}
{
a
n
}
{\displaystyle \{a_{n}\}}
a
1
=
2
,
a
n
+
1
−
a
n
=
a
n
−
a
n
+
1
a
n
2
a
n
+
1
2
(
n
∈
N
+
)
{\displaystyle a_{1}=2,a_{n+1}-a_{n}={\frac {a_{n}-a_{n+1}}{a_{n}^{2}a_{n+1}^{2}}}\quad (n\in \mathbb {N} ^{+})}
具有分式形式的遞推式有比較複雜的求解方法。但是有一類分式形式的遞推式實際上等價於很容易求解的周期數列，因此經常會出現在不需要給出詳細過程的選擇題或填空題中。
{
a
n
}
{\displaystyle \{a_{n}\}}
a
8
=
2
,
a
n
+
1
=
1
1
−
a
n
(
n
∈
N
+
)
{\displaystyle a_{8}=2,a_{n+1}={\frac {1}{1-a_{n}}}\quad (n\in \mathbb {N} ^{+})}
{
a
n
}
{\displaystyle \{a_{n}\}}
a
1
=
1
,
a
n
+
1
=
a
n
2
a
n
+
1
(
n
∈
N
+
)
{\displaystyle a_{1}=1,a_{n+1}={\frac {a_{n}}{2a_{n}+1}}\quad (n\in \mathbb {N} ^{+})}
{
a
n
}
{\displaystyle \{a_{n}\}}
a
1
=
0
,
a
n
+
1
=
a
n
−
3
3
a
n
+
1
(
n
∈
N
+
)
{\displaystyle a_{1}=0,a_{n+1}={\frac {a_{n}-{\sqrt {3}}}{{\sqrt {3}}a_{n}+1}}\quad (n\in \mathbb {N} ^{+})}
a
2018
{\displaystyle a_{2018}}
與集合、函數的關係
編輯
圖2-1-1 數列是一種自變量只取整數的離散化函數。 數列可以看成以正整數集
N
+
{\displaystyle \mathbb {N} ^{+}}
a
n
=
f
(
n
)
{\displaystyle a_{n}=f(n)}
y
=
f
(
x
)
{\displaystyle y=f(x)}
f
(
i
)
(
i
=
1
,
2
,
3
,
⋯
)
{\displaystyle f(i)\quad (i=1,2,3,\cdots )}
f
(
1
)
,
f
(
2
)
,
f
(
3
)
,
⋯
,
f
(
n
)
⋯
{\displaystyle f(1),f(2),f(3),\cdots ,f(n)\cdots }
常用結論與常見模型
編輯
根據規律猜想通項公式
編輯
2
3
,
4
5
,
6
7
,
8
9
,
.
.
.
{\displaystyle {\frac {2}{3}},{\frac {4}{5}},{\frac {6}{7}},{\frac {8}{9}},...}
16
17
{\displaystyle {\frac {16}{17}}}
18
19
{\displaystyle {\frac {18}{19}}}
20
21
{\displaystyle {\frac {20}{21}}}
22
23
{\displaystyle {\frac {22}{23}}}
0
,
1
4
,
1
3
,
3
8
,
.
.
.
,
n
−
1
2
n
,
.
.
.
{\displaystyle 0,{\frac {1}{4}},{\frac {1}{3}},{\frac {3}{8}},...,{\frac {n-1}{2n}},...}
3
7
{\displaystyle {\frac {3}{7}}}
1
2
,
3
4
,
5
6
,
7
8
,
.
.
.
{\displaystyle {\frac {1}{2}},{\frac {3}{4}},{\frac {5}{6}},{\frac {7}{8}},...}
2
,
5
,
2
2
,
11
,
.
.
.
{\displaystyle {\sqrt {2}},{\sqrt {5}},2{\sqrt {2}},{\sqrt {11}},...}
遞推式的簡單尋找
編輯
{
a
n
}
{\displaystyle \{a_{n}\}}
a
n
=
{
1
(
n
=
1
)
a
n
+
1
+
n
−
1
(
n
≥
2
,
n
∈
N
)
{\displaystyle a_{n}=\left\{{\begin{array}{l}1\quad (n=1)\\a_{n+1}+n-1\quad (n\geq 2,n\in \mathbb {N} )\end{array}}\right.}
a
n
=
{
1
(
n
=
1
)
a
n
−
1
+
n
(
n
≥
2
,
n
∈
N
)
{\displaystyle a_{n}=\left\{{\begin{array}{l}1\quad (n=1)\\a_{n-1}+n\quad (n\geq 2,n\in \mathbb {N} )\end{array}}\right.}
a
n
=
{
1
(
n
=
1
)
a
n
−
1
+
n
−
1
(
n
≥
2
,
n
∈
N
)
{\displaystyle a_{n}=\left\{{\begin{array}{l}1\quad (n=1)\\a_{n-1}+n-1\quad (n\geq 2,n\in \mathbb {N} )\end{array}}\right.}
a
n
=
{
1
(
n
=
1
)
a
n
+
1
+
n
+
1
(
n
≥
2
,
n
∈
N
)
{\displaystyle a_{n}=\left\{{\begin{array}{l}1\quad (n=1)\\a_{n+1}+n+1\quad (n\geq 2,n\in \mathbb {N} )\end{array}}\right.}
a
n
{\displaystyle a_{n}}
a
n
−
1
(
n
≥
2
)
{\displaystyle a_{n-1}\quad (n\geq 2)}
1
,
1
2
,
1
3
,
1
4
,
1
5
,
1
6
,
.
.
.
{\displaystyle 1,{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},{\frac {1}{5}},{\frac {1}{6}},...}
{
a
n
}
{\displaystyle \{a_{n}\}}
a
n
{\displaystyle a_{n}}
a
n
+
1
{\displaystyle a_{n+1}}
補充習題
編輯
(1)
1
2
,
3
4
,
7
8
,
15
16
,
31
32
,
.
.
.
{\displaystyle {\frac {1}{2}},{\frac {3}{4}},{\frac {7}{8}},{\frac {15}{16}},{\frac {31}{32}},...}
(2)
−
1
,
8
5
,
−
15
7
,
24
9
,
.
.
.
{\displaystyle -1,{\frac {8}{5}},-{\frac {15}{7}},{\frac {24}{9}},...}
參考資料
編輯
外部連結
編輯
相關例題1：
下列哪個數是數列 中的一項？( )
