(數學)三條AS/GS問題!!!急!!!

1(a) 設等比 = r
T4 = ar^3
T7 = ar^6
T4:T7 = ar^3 : ar^6 = 1/r^3 : 1
1/r^3 = 27
r = 1/3
無限項之和 = a + ar + ar^2 + ...
= a / (1 - r) = 3a/2
1(b) P(n)=2^(Tn)
P(1)*P(2)*P(3)*...
= (2^T1)(2^T2)(2^T3)...
= 2^[T1 + T2 + T3...]
= 2^(3a/2)
2(a) 月利率 = 0.06/12 = 0.005
一月後本利和 = x(1.005)
二月後本利和 = [x(1.005) + x](1.005) = x(1.005)^2 + x(1.005)
三月後本利和 = [x(1.005)^2 + x(1.005) + x](1.005) = x(1.005)^3 + x(1.005)^2 + x(1.005)
一年年尾的本利和 x(1.005) + x(1.005)^2 + x(1.005)^3 + ... + x(1.005)^12
= x(1.005)[(1.005)^12 - 1]/(1.005 - 1)
= $201x[(1.005)^12 - 1]
(b) (i) 他在第一年年尾的本利和 = $201(10000)[(1.005)^12 - 1] = $123,972.40
(ii) $2010000[(1.005)^n - 1] > 300000
(1.005)^n - 1 > 0.1493
(1.005)^n > 1.1493
nlog(1.005) > log(1.1493)
n > 27.9
要28個月才可儲得最少$300000
(c)如果該男子想在一年內儲$200000
$201x[(1.005)^12 - 1] = $200000
201(x)(0.06168) = 200000
x = $16,132.62
他每月需存入$16,132.62
3(a) 首n項的和是n(n + 2)
首n - 1項的和是(n - 1)(n - 1 + 2)
第n項 = n(n + 2) - (n - 1)(n + 1)
= n^2 + 2n - (n^2 - 1)
= 2n + 1
首項 = 2(1) + 1 = 3
公差 = 2
(b)i T(2) = 3 + 2 = 5
T(4) = 3 + 3(2) = 9
T(6) = 3 + 5(2) = 13
T(2),T(4),T(6),....,T(20)為一等差數列首項為5和公差為4,共10項。總和為
10[10 + 9(4)]/2 = 230
ii T(2) + T(4) + T(6) + T(8) = 4[10 + 3(4)]/2 = 44
T(10),T(12),T(14),....,T(20)的和 = 230 - 44 = 186