Question

在数列an中，a1=1,a(n+1)=an/(an+1)

在数列an中，a1=1,a(n+1)=an/(an+1)
1 求证(1/an)是等差数列，并求an通项公式
2设bn=1/(2^n乘an)，求数列bn的前n项和sn
详细过程

Answer 1

(1)
a(n+1)=an/(an+1)

1/a(n+1) = (an+1)/an
1/a(n+1) -1/an = 1
=>(1/an)是等差数列

1/an -1/a1= n-1
1/an =n
an =1/n
(2)
bn =1/(2^n.an)
= (1/2)[n(1/2)^(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=1/2
summation(i:1->n)i.(1/2)^(i-1)
=4[n.(1/2)^(n+1) - (n+1).(1/2)^n + 1]
=4(1- (n+2). (1/2)^(n+1) )
Sn = b1+b2+...+bn
= (1/2). {summation(i:1->n)i.(1/2)^(i-1)}
= 2(1- (n+2). (1/2)^(n+1) )

Answer 2

a(n+1)=an/(an+1)

二边取倒数得到:1/a(n+1)=1/an+1
即有1/a(n+1)-1/an=1
即数列{1/an}是一个首项是1/a1=1,公差是1的等差数列.
故有1/an=1+n-1=n
an=1/n
2,bn=1/(2^n)*n
Sn=1/2*1+1/(2^2)*2+1/2^3*3+...+1/2^n*n
1/2Sn=1/2^2*1+1/2^3*2+1/2^4*3+...+1/2^(n+1)*n
Sn-1/2Sn=1/2+1/2^2+1/2^3+1/2^n-1/2^(n+1)*n
1/2Sn=1/2*(1-1/2^n)/(1-1/2)-1/2^(n+1)*n
即有Sn=2-2/2^n-n/2^n

Answer 3

∵a(n+1)=an/(an+1)
两边取倒数：
∴1/a(n+1)=1+1/an
∴1/(a(n+1)-1/an=1
∴{1/an}是等差数列，公差为1
又a1=1
∴1/an=1/a1+(n-1)=n
∴an=1/n
(2)
bn=n/2^n
Sn=1/2+2/4+3/8+......+n/2^n
1/2Sn=1/4+2/8+3/16+...+(n-1)/2^n+n/2^(n+1)
相减：
1/2Sn=1/2+1/4+1/8+.......+1/2^n-n/2^(n+1)
=1/2[1-(1/2)^n]/(1-1/2)-n/2^(n+1)
=1-(n+2)/2^(n+1)
∴Sn=2-(n+2)/2^n

Answer 4

1、an+1=an/an+1，故1/an+1=an+1/an=1+1/an，即1/an+1-1/an=1
同理1/an-1/an-1=1,1/an-1-1/an-2=1，……
叠加得1/an=n