在数列{an}中，a1=1,a(n+1)=2an+2^n

b = a/2^(n-2)bn - b = an/2^(n-1) - a/2^(n-2)= (an - 2a )/2^(n-1)把 已知条件 a = 2an+2^n 即 an = 2a + 2^(n-1) 代入上式bn - b = 2^(n-1)/2^(n-1)= 1因此 bn 是等差数列b1 = a1/2^(1-1) = 1/1 = 1bn = n
2、
a(n+1)=2an+2^n
同除以2^n
a(n+1)/2^n=2an/2^n+1

a(n+1)/2^n-an/2^(n-1)=1
所以数列｛an/2^(n-1)｝为以1为公差的等差数列
a1/2^0=1
an/2^(n-1)=1+(n-1)*1=n

所以an = n2^(n-1)

Sn=1*2^0+2*2^1+3*2^2+....+ n2^(n-1)
2Sn= 1*2^1+2*2^2+....+(n-1)2^(n-1)+n2^n
用2式－1式
Sn=-1-2^1-2^2-....2^(n-1)+n2^n
=-1-(2+2^2+2^3+...+2^(n-1))+n2^n
=(n-1)2^n+1
网上找的，不知对不对