Question

已知数列{an},a1=1,an=2an-1+2的-n次方

问题如图

Answer 1

(1)
an= 2a(n-1) +2^(-n)
an/2^n - a(n-1)/2^(n-1) = 2^(-2n)
bn = a(n+1)/2^(n+1) - an/2^n 是等比数列

(2)
an/2^n - a(n-1)/2^(n-1) = 2^(-2n)
an/2^n - a1/2 = 2^(-4)+2^(-6)+...+2^(-2n)
= 2^(-4)( 1- 2^(-2(n-1)) / ( 1- 2^(-2))
= (1/12)( 1- 2^(-2(n-1))
an/2^n = 7/12 + (1/3)2^(-2n)
an = (7/12).2^n + (1/3).2^(-n)
Sn = a1+a2+...+an
= (7/6)(2^n-1) + (1/3) (1-2^(-n) )
= (7/6).2^n - (1/3).2^(-n) -5/6

追问

an/2^n-a(n-1)/2^(n-1)=2^(-2n) 怎么来的

an/2^n-a(n-1)/2^(n-1)=2^(-2n) 怎么来的

追答

an= 2a(n-1) +2^(-n)

an/2^n = [2a(n-1) +2^(-n)]/2^n
an/2^n - a(n-1)/2^)(n-1) = 2^(-2n)