新修订的高中数列题已知等差数列{An}的公差d不为0.设Sn=A1+A2q +……+Anq^n-1,Tn=A1-A2q+

1.
q=1
S3=A1+A2+A3=3A1+3d=15
A1=1 d=4
An=4n-3
2.
A1=d A2=2d A3=3d
S1=d
S2=d+2dq
S3=d+2dq+3dq^2
(d+2dq)^2=d(d+2dq+3dq^2)
d≠0
(1+2q)^2=1+2q+3q^2
1+4q+4q^2=1+2q+3q^2
q^2+2q=0
q=0(舍去) q=-2
3.
S2n=A1+A2q+……+A2nq^(2n-1)
qS2n=A1q+A2q^2+……+A2nq^2n
(1-q)S2n=A1+[(A2-A1)q+(A3-A2)q^2+……+(A2n-A(2n-1))q^(2n-1)]-A2nq^2n
=A1+d[q+q^2+q^3+……+q^(2n-1)]-A2nq^2n
T2n=A1-A2q+……-A2nq^(2n-1)
qT2n=A1q-A2q^2+……-A2nq^2n
(1+q)T2n=A1+[-(A2-A1)q+(A3-A2)q^2+……-(A2n-A(2n-1))q^(2n-1)]-A2nq^2n
=A1+d[-q+q^2-q^3+……-q^(2n-1)]-A2nq^2n
(1-q)S2n-(1+q)T2n=d[2q+2q^3+2q^5+……+2q^(2n-1)]
=2d[q+q^3+q^5+……+q^(2n-1)] 等比数列首项为q,公比为q^2,共n项
=2d×q(1-(q^2)^n)/(1-(q^2))
=2dq(1-q^2n)/(1-q^2)

等差数列怎么会有q?

你的题目看的我好累 确定没错了吗