等比與等比數列問題

1.
T(1) = 第一項
T(2) = 第二項
...
T(n) = 第 n 項

首四項即是 T(1), T(2), T(3), T(4)
(a) 5, 14, 29, 50
(b) 4, 7, 10, 13

2.
要推斷通項, 其中一個方法是觀察. 觀察會變的數字, 找出它和項數 n 的關係.

(a) 1．2, 2．3, 3．4, 4．5,　．．．
=>1．(1+1), 2．(2+1), 3．(3+1), 4．(4+1),　．．．
=> T(n) = n．(n+1)

(b) a，a^2，a^3，a^4， ．．．
=>a^1，a^2，a^3，a^4， ．．．
明顯地每項都有a, 所以 T(n)同樣有a.
T(n) = a^n

3.
首三項
(i) -1, 1, -1
(ii) 1, -1, 1

兩個數列之間的規律分別 有兩種表達方法.
一, (ii) 是 (i) 的每項乘 -1.
二, (ii) 是 (i) 的項數向左或向右移的結果.

n為代數, 由1起代入n中後運算

1a) T(n) = 3n^2 +2
T(1) = 3(1)^2 + 2 = 3(1) + 2 = 5
T(2) = 3(2)^2 + 2 = 3(4) + 2 = 14
T(3) = 3(3)^2 + 2 = 3(9) + 2 = 29
T(4) = 3(4)^2 + 2 = 3(16) + 2 = 50

1b) T(n) = 3n + 1
T(1) = 3(1) + 1 = 4
T(2) = 3(2) + 1 = 7
T(3) = 3(3) + 1 = 10
T(4) = 3(4) + 1 = 13

2a) 1．2，2．3，3．4，4．5，　．．．
T(n) = n(n+1)

2b) a，a^2，a^3，a^4， ．．．
T(n) = a^(n)


3i) T(n) = (-1)^n
T(1) = (-1)^1 = -1
T(2) = (-1)^2 = 1
T(3) = (-1)^3 = -1

3ii) T(n) = (-1)^n+1
T(1) = (-1)^1 + 1 = -1 + 1 = 0
T(2) = (-1)^2 + 1 = 1 + 1 = 2
T(3) = (-1)^3 + 1 = -1 + 1 = 0

or
3ii) T(n) = (-1)^(n+1)
T(1) = (-1)^(1 + 1) = -1^2 = 1
T(2) = (-1)^(2 + 1) = -1^3 = -1
T(3) = (-1)^(3 + 1) = -1^4 = 1