等差数列{an}的公差不为零,a4=7,a1,a2,a5成等比数列,数列{Tn}=a2+a4+a8+a2n具体题目：等差数列{an}的公差不为零,a4=7,a1,a2,a5成等比数列,数列{Tn}满足条件Tn=a2+a4+a8+……+a2^n,则Tn=?

Question

等差数列{an}的公差不为零,a4=7,a1,a2,a5成等比数列,数列{Tn}=a2+a4+a8+a2n
具体题目：等差数列{an}的公差不为零,a4=7,a1,a2,a5成等比数列,数列{Tn}满足条件Tn=a2+a4+a8+……+a2^n,则Tn=?

Answer 1

设公差为d；
a2=a4-3d=7-3d;
a1=7-4d;
a5=7+d;
由于a1,a2,a5成等比数列,a1*a5=a2*a2;
即(7-3d)*(7+d)=(7-2d)*(7-2d);
解方程得 d=0或7;
又因为公差不为0,所以d=7;
所以 an=7n-21;
再设bn=a2^n; bn=7*2^n-21;
所以Tn=7*(2+2^2+2^3+...+2^n)-21n;
Tn=14*(2^n-1)-21n=14*2^n-35n.