已知一等差數列2，10，18，.......的第n項是130。求..

T(1) = a = 2, d = 18 -10 = 10 -2 = 8

(a) T(n) = a + (n - 1)d
2 + 8(n - 1) = 130
8(n - 1) = 128
8n - 8 = 128
8n = 136
n = 17

(b) 2n = 17 * 2 = 34
S (n) = n / 2 [ 2a + (n - 1)d ]
S (34) = 34 / 2 [ 2 ( 2 ) + 8 ( 34 -1 )]
= 17 [ 4 + 264]
= 4556

n=17
b)4556

a) f(n) = 8n-6 = 130
8n = 136
n = 17

b) f(34) = 8(34)-6 = 266
首2n項之和 = 2+10+18+......+266 = 2x34 + 8x561 = 4556

2006-10-15 11:32:51 補充：
唔記得項數之和的公式, 現在以心算代替, 你自己查番條式喇

2006-10-15 11:35:13 補充：
樓上應為 T(n)=a (n-1)d

設a為首項,d為兩數的差
a.
等差數列式:T(n)=a+(n+1)d
即:130=2+(n+1)8
130=2+8n+8
120=8n
n=15

b. 2n=30
T(2n)=2+(30+1)8
T(2n)=250