已知数列{an}满足a1=b(b是常数),an=2an-1-2^n-1(n=2,3…)(1)证明:数列{an/2^n}是等差数列;(2)求数列{an}的通项公式;(3)求数列{an}的前n项和Sn

Question

已知数列{an}满足a1=b(b是常数),an=2an-1-2^n-1(n=2,3…)
(1)证明:数列{an/2^n}是等差数列;
(2)求数列{an}的通项公式;
(3)求数列{an}的前n项和Sn

Answer 1

1、an=2a(n-1)-2^(n-1)
an-2a(n-1)=-2^n-1
同时除以2^n
an/2^n-2a(n-1)/2^n=-1/2
an/2^n-a(n-1)/2^(n-1)=-1/2
即数列{an/2^n}是d=-1/2的等差数列
2、an/2^n=a1+(n-1)d=b-(n-1)/2
通项an=b2^n-2^n*(n-1)/2=b*2^n-(n-1)*2^(n-1) [n≥2]
a1=b
3、设Tn=1*2+2*2^2+3*2^3+4*2^4+.+n*2^n
2Tn=1*2^2+2*2^3+3*2^4+4*2^5+.+(n-1)*2^n+n*2^(n+1)
Tn-2Tn=2+2^2+2^3+2^4+.+*2^n-n*2^(n+1)=-2*(1-2^n)-n*2^(n+1)
Tn=2*(1-2^n)+n*2^(n+1)
∑(n-1)2^(n-1)=T(n-1)=2*[1-2^(n-1)]+(n-1)*2^n=2-2^(n+1)+n*2^n
Sn=b∑2^n-∑(n-1)2^(n-1)
=b*[-2(1-2^n)]-[2-2^(n+1)+n*2^n]
=b*(2*2^n-2)-[2-2^(n+1)+n*2^n] (n≥2)
S1=a1=b n=1