已知数列an是首项为a且公比q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12*S3、S6、S12-S6成等比数列2、求和：Tn=a1+2a4+3a7+……+na3n-2
已知数列an是首项为a且公比q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12*S3、S6、S12-S6成等比数列
2、求和：Tn=a1+2a4+3a7+……+na3n-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