问题 解答题
设数列{an}的前n项和为Sn,且Sn=2n-1.数列{bn}满足b1=2,bn+1-2bn=8an
(1)求数列{an}的通项公式;
(2)证明:数列{
bn
2n
}为等差数列,并求{bn}的前n项和Tn
答案

(1)当n=1时,a1=s1=21-1=1;

当n≥2时,an=Sn-Sn-1=(2n-1)-(2n-1-1)=2n-1

a1=1适合通项公式an=2n-1

∴an=2n-1(n∈N*);

(2)∵bn+1-2bn=8an

∴bn+1-2bn=2n+2

bn+1
2n+1
-
bn
2n
=2,又
b1
21
=1,

∴{

bn
2n
}是首项为1,公差为2的等等差数列.

bn
2n
=1+2(n-1)=2n-1,

∴bn=(2n-1)×2n

∴Tn=1×2+3×22+5×23+…+(2n-1)×2n

∴2Tn=1×22+3×23+…+(2n-3)×2n+(2n-1)×2n+1

∴-Tn=2+2(22+23+…+2n)-(2n-1)×2n+1

=2+2×

22(1-2n-1)
1-2
-(2n-1)×2n+1

=2n+2-6-(2n-1)×2n+1

=(3-2n)•2n+1-6,

∴Tn=(2n-3)•2n+1+6.

单项选择题 A1/A2型题
单项选择题