Question

已知数列an满足a1=1,an+1=(2^n+1an/(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；（3）设bn=n(n+1)an,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