数学数列问题 要过程

已知数列{an}是首相为a且公比不等于1的等比数列，Sn是其前n项和，a1,2a7,3a4成等差数列。

（1）证明12S3，S6,S12-S6成等比数列

（2）求和Tn=a1+2a4+3a7+...+na3n-2

A(n) = aq^(n-1), n = 1,2,...
S(n) = a[q^n - 1]/(q-1).

A(1) = a
2A(7) = 2aq^6,
3A(4) = 3aq^3,
2*2aq^6 = 2*2A(7) = A(1) + 3A(4) = a + 3aq^3,
0 = 4q^6 - 3q^3 - 1 = (4q^3 + 1)(q^3 - 1),
4q^3 + 1 = 0,
q^3 = -1/4.

12S(3) = 12*a[q^3 - 1]/[q-1] = 12a[-1/4 - 1]/[q-1] = 15a/[1-q]
S(6) = a[q^6 - 1]/[q-1] = a[1/16 - 1]/[q-1] = 15a/[16(1-q)]
S(12) - S(6) = a[q^12 - q^6]/[q-1] = a[1/16][1/16 - 1]/[1-q] = 15a/[(16)^2(q-1)]

12S(3)*[S(12) - S(6)] = (15a)^2/[16(q-1)]^2 = {15a/[16(1-q)]}^2 = [S(6)]^2
所以，12S3，S6,S12-S6成等比数列。哈，是等比数列哈。。

A(3n-2) = aq^(3n-3) = a(q^3)^(n-1) = a(-1/4)^(n-1)
T(n) = a + 2*a(-1/4) + 3*a(-1/4)^2 + ... + (n-1)*a(-1/4)^(n-2) + n*a(-1/4)^(n-1)

(-1/4)T(n) = 1*a(-1/4) + 2*a(-1/4)^2 + 3*a(-1/4)^3 + ... + (n-1)*a(-1/4)^(n-1) + n*a(-1/4)^n

T(n) - (-1/4)T(n) = a + a(-1/4) + a(-1/4)^2 + ... + a(-1/4)^(n-1) - n*a(-1/4)^n = a[1 - (-1/4)^n]/[1 - (-1/4)] - n*a(-1/4)^n
= 4a[1 - (-1/4)^n]/5 - na(-1/4)^n,

T(n) = {4a[1 - (-1/4)^n]/5 - na(-1/4)^n}*[1/(1+1/4)]
= {4a[1 - (-1/4)^n]/5 - na(-1/4)^n}*4/5
= 16a[1 - (-1/4)^n]/25 - 4na(-1/4)^n/5