已知{an}是公差不为零的等差数列,a1=1,a1,a3,a9成等比数列,令bn=2^an+n,求数列{bn}的前n项和sn

2025-06-22 18:01:44
推荐回答(2个)
回答1:

a3=1+2d
a9=1+8d
a1,a3,a9成等比数列
(1+2d)^2=1*(1+8d)
4d^2+4d+1=8d+1
4d^2-4d=0
d(d-1)=0
公差不为零
d=1
an=n
bn=2^n+n
Sn=2^(n+1)-1+n(n+1)/2

回答2:

设公差为c
1+8c=(1+2c)^2
c=1
sn=2+2^2+2^3+....+2^n+1+2+3+...n
sn=[2(1-2^n)/1-2]+(1+n)n/2=2^(n+1)-2+(1+n)n/2