# S5MathsCH1&2Notes 等差與等比數列

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S5MathsCH1&2Notes 等差與等比數列

(i) 0, 1, 2, 3, 4, … (ii) 2, 4, 6, 8, 10, … (iii) –1, 2, –4, 8, –16, … (iv) 1, 1, 2, 3, 5, …

(a) ∵ T(1) = 1 = (–1)1+1 T(2) = –1 = (–1)2+1 T(3) = 1 = (–1)3+1 T(4) = –1 = (–1)4+1 ∴ T(n) = (–1)n+1 (b) T(5) = (–1)5+1 = 1 T(8) = (–1)8+1 = –1 ∴ 該數列的第 5 項是 1，而第 8 項是 –1 。

T(1) = a T(2) = T(1) + d = a + d T(3) = T(2) + d = a + 2d T(4) = T(3) + d = a + 3d 因此，其通項為: T(n) = a + (n – 1)d

(a) 設公差是 d。 ∵ d = T(2) – T(1) = 1 – (–3) = 4 ∴ T(n) = –3 + (n – 1)(4) =

(b) 4k – 7 = 2009 4k = 2016 k =

–5, –5 + d, –5 + 2d, 7 ∵ 第 4 項亦可表示為 –5 + 3d。 ∴ –5 + 3d = 7 d = 4 ∴ 所求的兩個等差中項是 –1 和 3。

T(1) = a T(2) = T(1) × R = aR T(3) = T(2) × R = aR2 T(4) = T(3) × R = aR3 因此，其通項為: T(n) = aRn – 1，其中 R ≠ 0

(b) 若 T(k) = 1，求 k 的值。 (a) ∵ a = 128, R = 0.5 及 T(n) = aRn – 1 ∴ T(n) = 128(0.5)n – 1 = 27 × 21 – n =

(b) 若 T(k) = 1，求 k 的值。 (b) T(k) = 1 28 – k = 1 28 – k = 20 8 – k = 0 k =

= 101 = 101 = 101 這裏共有 50 對。 … + 100 = 50 × 101 高斯發現了一個快捷的方法來解這個問題。 = 5050

a + (a + d) + (a + 2d) + … + [a + (n – 1)d]， 我們以 l 表示該數列的末項， l = a + (n – 1)d

S(n) = a + (a + d) + (a + 2d) + … + (l – d) + l
S(n) = l (l – d) + (l – 2d) + … + (a + d) + a +) 2S(n) = (a + l) + (a + l) + (a + l) + … + (a + l) + (a + l) 即 2S(n) = n(a + l) ∵ l = a + (n – 1)d

a = 1, d = 2 – 1 = 1, l = 100

(c) 求第 11 項至第 20 項之和。 (a) ∵ d = 2 – 5 = –3 ∴ T(10) = 5 + (10 – 1)(–3) =

(b) S(10) = = (c) S(20) = = –470 ∴ T(11) + T(12) + … + T(20) = S(20) – S(10) = –470 – (–85) =

a + aR + aR2 + … + aRn – 1, 考慮下列兩個等比級數之和: S(n) = a + aR + aR2 + … + aRn – ……(1) RS(n) = aR + aR2 + … + aRn – 1 + aRn ……(2)

(1) – (2)： S(n) – RS(n) = a – aRn (1 – R)S(n) = a(1 – Rn) 其中 a 是首項，R 是公比及 n 是項數。 當 R < 1 時，利用此公式會較為方便。

∴ 此公式亦可寫成: 當 R > 1 時，利用此公式會較為方便。

∵ a = 3125，R = 及 T(n) = aRn – 1 = 1

∴ n – 1 = 5 n = 6 ∴ S(6) = =

Font size 36, bold, theme color of the chapter (red for geometry, blue for algebra, green for statistics)

Rn  0；當 n  ， 一個等比級數的無限項之和為: