设{an}是一个公差不为0的等差数列,它的前十项和S10为110,且a1,a2,a4成等比数列.

1.a1
a2=a1+d
a4=a1+3d
a2^2=a1*a4
(a1+d)^2=a1*(a1+3d)
2a1*d+d^2=3a1*d
d=a1
S10=110
=10a1+10*9*d/2
a1=d=2
{an}=2n
2.Tn=1*2^1+2*2^2+...+(n-1)*2^(n-1)+n*2^n
2Tn=1*2^2+2*2^3+...+(n-1)*2^n+n*2^(n+1)
Tn-2Tn=-Tn=1*2^1+1*2^2+1*2^3+...+1*2^n-n*2^(n+1)
=(2^1+2^2+2^3+...+2^n)-n*2^(n+1)
=(1-n)*2^(n+1)-2
Tn=2+(n-1)*2^(n+1)

1.由已知有s10=10a1+45d=110
a1(a1+3d)=(a1+d)^2
解得 a1=d=2 故an=2n
2.Bn=n*2^n
Tn=1*2^1+2*2^2+……+n*2^n
2Tn=1*2^2+2*2^3+……+(n-1)*2^n+n*2^(n+1)
上两式相减得Tn=(n+1)*2^(n+1)-2