数列{an}满足a1＝1,an＋1＝2^n＋1*an/an＋2^n(n∈N+)1)证明：数列{2^n/an}是等差数列2)求数列{an}的通项公式an3)设bn＝n(n+1)a,求数列{bn}的前n项和Sn

Question

数列{an}满足a1＝1,an＋1＝2^n＋1*an/an＋2^n (n∈N+)
1)证明：数列{2^n/an}是等差数列
2)求数列{an}的通项公式an
3)设bn＝n(n+1)a,求数列{bn}的前n项和Sn

Answer 1

a(n+1)=2^(n+1)an/[an+2^n] 等式两边同时除以2^(n+1)
a(n+1)/2^(n+1)=2^(n+1)an/[2^(n+1)(an+2^n)]
a(n+1)/2^(n+1)=an/(an+2^n)]取倒数
2^(n+1)/a(n+1)=(an+2^n)/an
2^(n+1)/a(n+1)=2^n/an+1
2^(n+1)/a(n+1)-2^n/an=1
所以数列{2^n/an}是以1为公差的等差数列
2^n/an=2^1/a1+n-1
2^n/an=2+n-1
2^n/an=n+1取倒数
an=2^n/(n+1)
bn=n(n+1)an
=2^n/(n+1)*n(n+1)
=n*2^n
sn=1*2^1+2*2^2+3*2^3+.+n*2^n
2sn=1*2^2+2*2^3+3*2^4+.+(n-1)*2^n+n*2^(n+1)
sn-2sn=2^1+2^2+2^3+.+2^n-n*2^(n+1)
-sn=2*(1-2^n)/(1-2)-n*2^(n+1)
-sn=2^(n+1)-2-n*2^(n+1)
sn=n*2^(n+1)-2^(n+1)+2
=(n-1)*2^(n+1)+2