the general term

4(n-1) 不等於1 !!!

2010-06-06 20:18:42 補充：
http://hk.knowledge.yahoo.com/question/question?qid=7009060900762

2010-06-06 20:21:52 補充：
Let An=an^3+bn^2+cn+d
A1 = a + b + c + d = 1 ...(1)
A2 = 8a +
4b + 2c + d = 4 ...(2)
A3 = 27a + 9b + 3c + d = 8 ... (3)
A4 =
64a + 16b + 4c + d= 12 ... (4)

(1)-(2)=>
7a + 3b + c = 3
... (5)
(3)-(2)=>
19a + 5b + c = 4 ... (6)
(4)-(3)=>
37a
+ 7b + c = 4 ... (7)

(6)-(5) =>
12a + 2b = 1 ... (8)
(7)-(6)
=>
18a + 2b = 0 ... (9)

(9)-(8) =>
6a = -1
a =
-1/6

Sub into (9),
-3+2b =0
b= 3/2

Sub into (7),
-7/6
+ 9/2 + c = 3
c=-1/3

Sub into (1),
-1/6 + 3/2 - 1/3 + d = 1
d=0

So
An=-1/6n^3+3/2n^2-n/3
An = -n(n^2-9n+2)/6

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Differences:
...1...4...8...12
......3...4...4
.........2...0
............2

其實根本唔能夠確定佢係幾次(highest power)，
因為唔夠data

2010-06-07 19:13:30 補充：
但只可以let做3次...(3次以下明顯冇可能)。
假如let 做n次(n>3,n belongs to R)，
咁就要n 個given terms，
因為deg(possible closed form)=n => possible closed form 有n 個variables，
(姐係要n條方程)
所以無解

基於以上原因，Vlc既答案相信已是最準確

你問乜呀?!你自己3寒月 8左呢個係正方形數列啦...佢係咁加4炸嘛...

The General Term of the sequence 1,4,8,12.... is?
T(n) : 4(n-1)

如果想知道此數如可計算,我可以以高斯求和法教你