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{
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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
=2,又bn 2n
=1,b1 21
∴{
}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
∴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×
-(2n-1)×2n+122(1-2n-1) 1-2
=2n+2-6-(2n-1)×2n+1,
=(3-2n)•2n+1-6,
∴Tn=(2n-3)•2n+1+6.