已知数列{an}满足a1=b(b是常数),an=2an-1-2^n-1(n=2,3…)
问题描述：
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
答
1、an=2a(n-1)-2^(n-1)an-2a(n-1)=-2^n-1同时除以2^nan/2^n-2a(n-1)/2^n=-1/2an/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-...