Question

在数列an中，a1=1，an+1 2an+2的n次方

Answer 1

（1）证明：
a(n+1)=2an+2^n
bn=an/2^(n-1)
b(n+1)=a(n+1)/2^n=2an/2^n+1
b(n+1)-bn
=(2an/2^n+1)-(an/2^(n-1))
=(2an/(2*2^(n-1))+1)-(an/2^(n-1))
=(an/2^(n-1)+1)-(an/2^(n-1))
=an/2^(n-1)+1-an/2^(n-1)
=1
∴{bn}是等差数列
（2）解：
b1=a1/2^(1-1)=1/2^0=1/1=1
bn的公差为1
∴bn=n
∴an=n2^(n-1)
sn=1*2^0+2*2^1+3*2^2+...+n2^(n-1)
2sn=1*2^1+2*2^2+3*2^3+...+n2^n
(2-1)sn
=sn
=-1*2^0-1*2^1-1*2^2-...-1*2^(n-1)+n2^n
=-(2^0+2^1+...+2^(n-1))+n2^n
=-(1-2^n)/(1-2)+n2^n
=1-2^n+n*2^n
=(n-1)2^n+1