Question

已知数列｛an}是公差不为零的等差数列且a1=1，且a2，a4，a8成等比数列。①求通项an②令bn=an2^an求数列bn

求bn前n项和Sn

Answer 1

1、an=a1+(n-1)d
a4²=a2*a8
(1+3d)²=(1+d)(1+7d)
1+6d+9d²=1+8d+7d²
d≠0
∴d=1
an=n
2、 bn=n2^n
Sn=1*2+2*2²+3*2³+……+n2^n
2Sn= 1*2²+2*2³+……+(n-1)2^n+n2^(n+1)
2Sn-Sn=1*2²+2*2³+……+(n-1)2^n+n2^(n+1) - （1*2+2*2²+3*2³+……+n2^n ）
Sn= -2-2²-2³-……-2^n+n2^(n+1)
=n2^(n+1)-(2+2²+2³+……+2^n)
等比数列 2+2²+2³+……+2^n=2(1-2^n)/1-2=2(2^n-1)
∴Sn=n2^(n+1)-2(2^n-1)=(n-1)2^(n+1)+2

Answer 2

a4²=a2*a8
(a1+3d)²=(a1+d)(a1+7d)
a1²+6a1d+9d²=a1²+8a1d+7d² a1=1 d≠0
∴d=1
通项an=1+(n-1)=n
bn=n2^n