数列{An}满足A（n+1）=2An+2^n,A1=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

(错位相消法)