a1=1,a(n+1)=2an+2^n

等式两边同除以2^(n+1)得

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

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

{an/2^n}是公差为1/2等差数列

an/2^n=a1/2+1/2(n-1)=n/2

an=n*2^(n-1)

a2=2*2^(2-1)=4,a3=3*2^(3-1)=12,a4=4*2^(4-1)=32

a2=4,a3=12,a4=32

bn=an/2^(n-1)=n*2^(n-1)/2^(n-1)=n

bn=n,b(n-1)=n-1

bn-b(n-1)=n-(n-1)=1

即{bn}是公差为1,等差数列

3)Sn=1+2*2+3*2^2+....+(n-1)*2^(n-2)+n*2^(n-1)

2Sn= 2+2*2^2+3*2^3+....+(n-1)*2^(n-1)+n*2^n

两式相减得

-Sn=1+2+2^2+....+2^(n-1)-n*2^n=(1-2^n)/(1-2)-n*2^n=-(1-2^n)-n*2^n

Sn=(1-2^n)+n*2^n=[(n-1)*2^n]-1

Sn=[(n-1)*2^n]-1