(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) )