Question

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

(1)求证{1/an}是等差数列，并求数列{an}的通项公式
(2)求数列{2^n/an}前n项和Sn

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

Answer 2

(1)
a(n+1)+an*a(n+1)-an=0

1= 1/a(n+1) -1/an
=>{1/an}是等差数列
1/a(n+1)- 1/an =1
1/an -1/a1= n-1
1/an = n
an =1/n
(2)
2^n/an = n.2^n = 2.(n.2^(n-1))
consider
1+x+x^2+..+x^n = (x^(n+1)-1)/(x-1)
1+2x+..+nx^(n-1) = [(x^(n+1)-1)/(x-1)]'
= [ nx^(n+1) -(n+1)x^n +1 ] /(x-1)^2
put x=2
1.2^0 + 2.2^1+...+n.2^(n-1)
= n.2^(n+1)- (n+1).2^n + 1
= (n-1).2^n + 1
Sn = 2[ (n-1).2^n + 1 ]
= [(n-1).2^(n+1)] + 2