maths!等差問題...急急急

1)設S(n)為等差級數
3,6,9...
i)求n使得 S(n) = 315
S(n) = 3 + 3(n - 1) = 315
3(n - 1) = 312
n = 105

ii)求n的最大值使得S(n) < 600
3 + 3(n - 1) < 600
3(n - 1) < 597
n < 200
最大值 = 199

2)己知等差數列的第二項為第五項的九倍和數列首八項的和為56,求數列的首項和公差
T(n) = a + (n - 1) d
T(2) = a + d
T(5) = a + 4d
a + d = 9(a + 4d)
8a = -35d ----------- (1)
S(n) = (n/2) [ 2a + (n - 1) d ]
S(8) = (4) [ 2a + 7d ] = 56
2a + 7d = 14 ---------- (2)
solving (1) and (2), d = -8, a = 35

(i)
S(n)=(n/2)[2(3) + (n - 1)3]=315
..................n(6 + 3n - 3)=630
.............3n^2 + 3n - 630=0
.................n^2 + n - 210=0
...............(n - 14)(n + 15)=0
..................................n=14或-15

(ii)
S(n)=(n/2)[2(3) + (n - 1)3&lt;600
..................n(6 + 3n - 3)&lt;1200
.............3n^2 + 3n - 1200&lt;0
.................n^2 + n - 400&lt;0
..................................n&lt;19.5或n&lt;-20.5
所以-20&lt;20

(2)
設首項為a和公差為d
第二項=a + (2 - 1)d=a + d
第五項=a + (5 - 1)d=a + 4d
首八項的和=(8/2)[2a + (8 - 1)d]=8a + 28d

第二項=(第五項)
a + d=9(a + 4d)
a + d=9a + 36d
8a + 35d=0-------------------{1}
首八項的和=8a + 28d=56----------{2}
{1} - 4{2} :
..8a + 35d=0
- 8a + 28d=56
...........7d=-56
............d=-8
代d=-8入{2}
8a + 28(-8)=56
..............a=35

2007-09-16 00:28:29 補充：
我想問你個S(n)係指&quot;首n項ge和&quot;定&quot;首n項&quot;如果係首n項ge和...就係我上面計ge如果係首n項就係:(i)S(n)=3 (n - 1)3]=315......................3n=315........................n=105(ii)S(n)=3 (n - 1)3

2007-09-16 00:29:12 補充：
(ii)S(n)=3 (n - 1)3

2007-09-16 00:29:59 補充：
(ii)S(n)=3 + (n - 1)3&lt;600....................3n&lt;600......................n&lt;200