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# (40)等差級數 請看以下的數列

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(1)

## 40-1

### (40)等差級數

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5 𝑎6 𝑎7 𝑎8 𝑎9 𝑎10

1 3 5 7 9 11 13 15 17 19

𝑎𝑖+1= 𝑎𝑖+ 𝑑 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (1) (1) 𝑎1= 2，𝑑 = 3

𝑎2 = 2 + 3 = 5 𝑎3 = 5 + 3 = 8 𝑎4 = 8 + 3 = 11 𝑎5 = 11 + 3 = 14

3，5，8，11，14 是一個等差級數

(2) 𝑎1= 1，𝑑 = −2 𝑎2 = 1 − 2 = −1 𝑎3 = −1 − 2 = −3 𝑎4 = −3 − 2 = −5 𝑎5 = −5 − 2 = −7

∴ 1，-1，-3，-5，-7 是一等差級數

(2)

## 40-2

(3) 𝑎1= 0，𝑑 =1

2

𝑎2 = 0 +1 2=1

2 𝑎3 =1

2+1 2= 1 𝑎4 = 1 +1

2=3 2 𝑎5 =3

2+1 2= 2 0，1

2，1，3

2，2是一等差級數

(4) 𝑎1= 1，𝑑 = −1

2

𝑎2 = 1 −1 2=1

2 𝑎3 =1

2−1 2= 0 𝑎4 = 0 −1

2= −1 2 𝑎5 = −1

2−1

2= −1 𝑎6 = −1 −1

2= −3 2 𝑎7 = −3

2−1

2= −2

∴ 1，1

2，0， −1

2， −3

2， − 2是一等差級數

𝑎2 = 𝑎1+ 𝑑

𝑎3 = 𝑎2+ 𝑑 = 𝑎1+ 2𝑑 𝑎4 = 𝑎3+ 𝑑 = 𝑎1+ 3𝑑 由此我們可以得到

𝑎𝑖 = 𝑎1+ (𝑖 − 1)𝑑 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (2)

(3)

## 40-3

(5) 𝑎1= 2，𝑑 = 3

𝑎4 = 2 + (4 − 1) × 3 = 2 + 9 = 11 我們也可以計算一下來驗證

𝑎1 = 2

𝑎2 = 2 + 3 = 5

𝑎3 = 𝑎2+ 3 = 5 + 3 = 8 𝑎4 = 𝑎3+ 3 = 8 + 3 = 11 (6) 𝑎1= −1，𝑑 = −2

𝑎5 = −1 + (5 − 1) × (−2) = −1 + 4 × (−2) = −1 − 8 = −9 驗證

𝑎1 = −1

𝑎2 = −1 − 2 = −3 𝑎3 = −3 − 2 = −5 𝑎4 = −5 − 2 = −7 𝑎5 = −7 − 2 = −9

(7) 𝑎1= 1，𝑑 =1

2

𝑎4 = 1 + (4 − 1) × (1

2) = 1 + 3 × (1

2) = 1 +3 2= 5

2 驗證

𝑎1 = 1 𝑎2 = 1 +1

2=3 2 𝑎3 = 3

2+1 2= 2 𝑎4 = 2 +1

2=5 2

(4)

## 40-4

(8) 𝑎1= −2，𝑑 = −1

2

𝑎3 = −2 + (3 − 1) (−1

2) = −2 + 2 (−1

2) = −2 − 1 = −3 驗證

𝑎1 = −2 𝑎2 = −2 −1

2= −5 2 𝑎3 = −5

2−1 2= −6

2= −3

(9) 𝑎𝑖 = 2i + 1 這個數列如下:

𝑎1= 2(1) + 1 = 3 𝑎2 = 2(2) + 1 = 5 𝑎3 = 2(3) + 1 = 7 𝑎4 = 2(4) + 1 = 9 𝑎5 = 2(5) + 1 = 11 以下是這個數列:

3，5，7，9，11，13，15，17，19，21

∴ 𝑎1 = 2(1) + 1 = 3

𝑑 = 𝑎𝑖 − 𝑎𝑖−1= (2𝑖 + 1) − [2(𝑖 − 1) + 1] = 2𝑖 + 1 − 2𝑖 + 2 − 1 = 2

(5)

## 40-5

(10) 𝑎𝑖 = −2i + 1 𝑎1= −2(1) + 1 = −1 𝑎2 = −2(2) + 1 = −3 𝑎3 = −2(3) + 1 = −5 𝑎4 = −2(4) + 1 = −7 以下是這個數列:

-1，-3，-5，-7，-9，-11，-13 這也是一個等差級數

𝑎1 = −2(1) + 1 = −2 + 1 = −1

𝑑 = 𝑎𝑖 − 𝑎𝑖−1= (−2𝑖 + 1) − [−2(𝑖 − 1) + 1] = −2𝑖 + 1 + 2𝑖 − 2 − 1 = −2 等差級數的和

∑ 𝑎𝑖

𝑛

𝑖=1

= 𝑎1+ 𝑎2+ ⋯ + 𝑎𝑛

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5 𝑎6 𝑎7 𝑎8

1 3 5 7 9 11 13 15

∑ 𝑎𝑖

𝑛

𝑖=1

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64

(6)

## 40-6

𝑎1+ 𝑎𝑛 = 𝑎2+ 𝑎𝑛−1= 𝑎3+ 𝑎𝑛−2= 𝑎4+ 𝑎𝑛−3 如果一個等差級數的首項是𝑎1，公差是 d，則

𝑎𝑖 = 𝑎1+ (𝑖 − 1)𝑑 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (3) 末項𝑎𝑛 = 𝑎1+ (𝑛 − 1)𝑑 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (4)

𝑎7 = 𝑎1+ (7 − 1)𝑑 = (−2) + 6(3) = −2 + 18 = 16

2，𝑛 = 5 𝑎5 = 𝑎1+ (5 − 1)𝑑 = 1 + 4 ×1

2= 1 + 2 = 3 假設𝑎1 = −1，𝑑 = −2，𝑛 = 4

𝑎4 = 𝑎1+ (4 − 1)𝑑 = −1 + 3(−2) = −1 − 6 = −7 以下的式子是很容易了解的:

𝑎1+ 𝑎𝑛 = 𝑎1+ (𝑎1+ (𝑛 − 1)𝑑) = 2𝑎1 + (𝑛 − 1)𝑑

𝑎2+ 𝑎𝑛−1= (𝑎1+ 𝑑) + (𝑎1+ (𝑛 − 2)𝑑) = 2𝑎1+ (𝑛 − 1)𝑑 𝑎3+ 𝑎𝑛−2= (𝑎1+ 2𝑑) + (𝑎1+ (𝑛 − 3)𝑑) = 2𝑎1+ (𝑛 − 1)𝑑 𝑎4+ 𝑎𝑛−3= (𝑎1+ 3𝑑) + (𝑎1+ (𝑛 − 4)𝑑) = 2𝑎1+ (𝑛 − 1)𝑑

2份

(7)

## 40-7

(𝑎1，𝑎𝑛) (𝑎2，𝑎𝑛−1) ⋮

(𝑎𝑛 2，𝑎𝑛

2+1)

∑ 𝑎𝑖

𝑛

𝑖=1

=𝑛

2(𝑎1+ 𝑎𝑛) ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (5)

2(𝑎1+ 𝑎𝑛)仍然是成立 的。

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5

1 3 5 7 9

∑ 𝑎𝑖

5

𝑖=1

= 1 + 3 + 5 + 7 + 9 = 25

𝑎2+ 𝑎𝑛−1= 𝑎2+ 𝑎4 = 3 + 7 = 10 但還有一個𝑎3 = 5，我們知道

𝑎3 = 𝑎1+ 𝑎𝑛

2 = 𝑎1+ 𝑎5

2 =1 + 9 2 = 5

(8)

## 40-8

2

= 𝑎1+𝑎𝑛

2

2

+ 𝑎𝑛+3

2

∴ ∑ 𝑎𝑖

𝑛

𝑖=1

= (𝑎1+ 𝑎𝑛)𝑛 − 1

2 +𝑎1+ 𝑎𝑛

2 = (𝑎1+ 𝑎𝑛)(𝑛 − 1 2 +1

2) = (𝑎1+ 𝑎𝑛)𝑛 2

∑ 𝑎𝑖

𝑛

𝑖=1

= (𝑎1+ 𝑎𝑛

2 )𝑛 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (6)

2 一定是偶數嗎?

∴ 𝑎1+ 𝑎𝑛 = 𝑎1+ (𝑎1+ (𝑛 − 1)𝑑) = 2𝑎1+ (𝑛 − 1)𝑑 但(𝑛 − 1)是偶數

∴ 𝑎1+ 𝑎𝑛 = 2𝑎1+ 2ℎ𝑑

∴ 𝑎1+ 𝑎𝑛是偶數 (11)

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5 𝑎6

1 3 5 7 9 11

𝑛 = 6(偶數)

∑ 𝑎𝑖

6

𝑖=1

= (𝑎1+ 𝑎𝑛)𝑛

2= (1 + 11)6

2= (12)(3) = 36

(9)

## 40-9

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5 𝑎6 𝑎7

1 3 5 7 9 11 13

𝑛 = 7(奇數)

∑ 𝑎𝑖

7

𝑖=1

= (𝑎1+ 𝑎𝑛

2 )𝑛 = 1 + 13

2 × 7 = 7 × 7 = 49

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5 𝑎6 -3 -5 -7 -9 -11 -13

𝑛 = 6(偶數)

∑ 𝑎𝑖

6

𝑖=1

= (𝑎1+ 𝑎6)6

2= (−3 − 13)6

2= (−16)6

2= −48

(14)

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5 𝑎6 𝑎7

-2 0 2 4 6 8 10

𝑛 = 7(奇數)

∑ 𝑎𝑖

7

𝑖=1

= (𝑎1+ 𝑎𝑛

2 )𝑛 = (−2 + 10

2 ) × 7 =8

2× 7 = 28

(10)

## 40-10

∑ 𝑎𝑖

𝑛

𝑖=1

=(𝑎1+ 𝑎𝑛)𝑛

2 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (7)

∵ 𝑎1 = 𝑎1，𝑎𝑛 = 𝑎1+ (𝑛 − 1)𝑑

∴ ∑ 𝑎𝑖

𝑛

𝑖=1

=𝑛

2(𝑎1+ 𝑎𝑛) =𝑛

2(𝑎1+ 𝑎1+ (𝑛 − 1)𝑑) =𝑛

2(2𝑎1+ (𝑛 − 1)𝑑)

= 𝑛𝑎1+𝑛(𝑛 − 1)

2 d ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (8) (15)

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5 𝑎6 -3 -5 -7 -9 -11 -13

𝑎1 = −3，𝑑 = −2，𝑛 = 6

∴ ∑ 𝑎𝑖

𝑛

𝑖=1

= 𝑛𝑎1+𝑛(𝑛 − 1)

2 d = 6(−3) +6 × 5

2 (−2) = −18 − 30 = −48

∑ 𝑎𝑖

6

𝑖=1

= (𝑎1+ 𝑎6)𝑛

2= (−3 − 13)6

2= (−16)3 = −48

(16)

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5 𝑎6 𝑎7

-2 0 2 4 6 8 10

𝑎1 = −2，𝑑 = 2，𝑛 = 7

(11)

## 40-11

∑ 𝑎𝑖

7

𝑖=1

= 𝑛𝑎1+𝑛(𝑛 − 1)

2 d = 7(−2) +7 × 6

2 (2) = −14 +42

2 (2) = 28

∑ 𝑎𝑖

7

𝑖=1

= (𝑎1+ 𝑎𝑛

2 )𝑛 = (−2 + 10

2 ) × 7 = 4 × 7 = 28

(17)

𝑎1 𝑎2 𝑎3 𝑎4 𝑎5

1 2 3 4 5

𝑎1 = 1，𝑑 = 1，𝑛 = 5

∑ 𝑎𝑖

5

𝑖=1

= 𝑛𝑎1+𝑛(𝑛 − 1)

2 d = 5 × 1 +5 × 4

2 × 1 = 5 + 10 = 15

∑ 𝑎𝑖

5

𝑖=1

= (𝑎1+ 𝑎𝑛

2 )𝑛 = (1 + 5

2 ) × 5 = 3 × 5 = 15

𝑎1 = 1，𝑑 = 1

∑ 𝑎𝑖

𝑛

𝑖=1

= 𝑎1+ 𝑎2+ ⋯ + 𝑎𝑛

∑ 𝑎𝑖

𝑛

𝑖=1

= 𝑛𝑎1+𝑛(𝑛 − 1)

2 d = n +𝑛(𝑛 − 1)

2 =2𝑛 + 𝑛(𝑛 − 1)

2 = 𝑛(𝑛 + 1) 2

(12)

## 40-12

∑ 𝑎𝑖

𝑛

𝑖=1

= (𝑎1+ 𝑎𝑛

2 )𝑛 = 1 + 𝑛

2 𝑛 =𝑛(𝑛 + 1) 2

∑ 𝑎𝑖

100

𝑖=1

=𝑛(𝑛 + 1)

2 =100 × 101

2 = 50 × 101 = 5050

∑ 𝑎𝑖

3

𝑖=1

=𝑛(𝑛 + 1)

2 =3 × 4 2 = 6

∑ 𝑎𝑖

5

𝑖=1

=𝑛(𝑛 + 1)

2 =5 × 6

2 = 15 我們現在繼續討論等差級數的和

∑ 𝑎𝑖

𝑛

𝑖=1

=𝑛

2(𝑎1+ 𝑎𝑛) =𝑛

2(𝑎1+ 𝑎1 + (𝑛 − 1)𝑑) =𝑛

2(2𝑎1+ (𝑛 − 1)𝑑)

=𝑑

2𝑛2+ (𝑎1−𝑑

2)𝑛 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (9) 我們可以利用公式(7)來計算等差級數的和

(18) 𝑎1 = 1，𝑑 = 2

∑ 𝑎𝑖

𝑛

𝑖=1

=𝑑

2𝑛2+ (𝑎1−𝑑

2)𝑛 =2

2𝑛2+ (1 −2

2)𝑛 = 𝑛2

∑ 𝑎𝑖

𝑛

𝑖=1

= (𝑎1+ 𝑎𝑛

2 )𝑛 = 1 + 11

2 × 6 = 6 × 6 = 36 = 𝑛2

(13)

## 40-13

(19) 𝑎1 = 1，𝑑 = −2

∑ 𝑎𝑖

𝑛

𝑖=1

=𝑑

2𝑛2+ (𝑎1−𝑑

2) 𝑛 =−2

2 𝑛2 + (1 +2

2) 𝑛 = −𝑛2 + 2𝑛

∑ 𝑎𝑖

𝑛

𝑖=1

= (𝑎1+ 𝑎𝑛

2 ) 𝑛 =1 − 11

2 × 7 =−10

2 × 7 = −35

−𝑛2+ 2𝑛 = −(7)2+ 2 × 7 = −49 + 14 = −35

𝑎𝑛2+ 𝑏𝑛 = 𝑑

2𝑛2+ (𝑎1−𝑑 2) 𝑛

𝑑

2= 𝑎 ∴ 𝑑 = 2𝑎

(𝑎1−𝑑

2) = 𝑏 ∴ 𝑎1 = 𝑑

2+ 𝑏 = 𝑎 + 𝑏

𝑑 = 2𝑎，𝑎1 = 𝑎 + 𝑏 ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ (10) (20) 𝑎𝑛2+ 𝑏𝑛 = 𝑛2

𝑎 = 1，𝑏 = 0

∴ 𝑑 = 2𝑎 = 2 𝑎1 = 𝑎 + 𝑏 = 1

(14)

## 40-14

(21) 𝑎𝑛2+ 𝑏𝑛 = −𝑛2+ 2𝑛 𝑎 = −1，𝑏 = 2

𝑑 = 2𝑎 = −2

𝑎1 = 𝑎 + 𝑏 = −1 + 2 = 1

[r]

Weierstrass 是用級數來研究與刻劃複變函數的性質, 在第一講與第二講中我們已提到全 純函數的冪 級數展開, 以及 Weierstrass M -

2.通過標準：國語文及外語文性向測驗成績兩科均達 平均數正 1.5 個標準差以上或 百分等級 93 以上，且其中一科須達平均數正 2

[r]

• 中年級 ：在數方面，能掌握自然數的四則與混合運算，並初