等差数列{an}公差d≠0,a1,a2,a5成等比数列,数列{Tn}满足Tn=a2+a4+a8+a(2的n次方),则Tn=
问题描述:
等差数列{an}公差d≠0,a1,a2,a5成等比数列,数列{Tn}满足Tn=a2+a4+a8+a(2的n次方),则Tn=
答
an等于2n-1
重新构造数列2的n+1次方减一,再求和就是Tn我算一会,等会打答案行么?答案是Tn=2的n+2次方减n减4你可以代进去试试!