已知数列an是首项为a 且公比q不等于一1的等比数列 sn是其前n项和 a1 2a7 3a4成等差数列

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已知数列an是首项为a 且公比q不等于一1的等比数列 sn是其前n项和 a1 2a7 3a4成等差数列
已证得12s3 s6 s12-s6成等比数列,求和 Tn=a1+2a4+3a7+…+na(3n-2)

a1,a7,a4成等差数列
2a7=a1+a4
2a1q^6=a1+a1q^3
2q^6=1+q^3
2q^6-q^3-1=(2q^3+1)(q^3-1)=0
因为公比Q不等于1,
所以,q^3=-1/2,
2S3*（S12-S6)
=2a1(1-q^3)/(1-q)*[a1(1-q^12)/(1-q)-a1(1-q^6)/(1-q)]
=2a1(1+1/2)/(1-q)*[a1(1-1/16)/(1-q)-a1(1-1/4)/(1-q)]
=[a1/(1-q)]^2[3*(15/16-3/4)
=[a1/(1-q)]^2*9/16
=[a1*(3/4)/(1-q)]^2
=[a1*(1-1/4)/(1-q)]^2
=[a1*(1-q^6)/(1-q)]^2
=S6^2
2S3,S6,S12-S6等比
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

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