Question

已知各项均为正数的数列{an}满足a1=1,an+1+an*an+1-an=0

Answer 1

a[n+1]+ana[n+1]-an=0

a[n+1]-an=-ana[n+1]

二边同除以ana[n+1]

1/an-1/a[n+1]=-1

即有1/a[n+1]-1/an=1

故{1/an}是一个首项是1/a1=1,公差是1的等差数列.

故1/an=1+1*(n-1)=n

an=1/n

2.bn=2^n/an=2^n*n

Sn=1*2+2*2^2+3*2^3+...+n*2^n

2Sn=1*2^2+2*2^3+3*2^4+...+n*2^(n+1)

Sn-2Sn=2+2^2+2^3+.+2^n-n*2^(n+1)

-Sn=2*(2^n-1)/(2-1)-2n*2^n

Sn=2(n-1)*2^n+2