a(n+1)=2an+2^n两边同时除以2^n+1a(n+1)/(2^n+1)=an/(2^n)-1a(n+1)/(2^n+1)-an/(2^n)=1所以｛an/(2^n)｝是以1/2为首项,1为公差的等差数列所以an/(2^n)=1/2+(n-1)*(1/2)=n/2所以an=(n/2)*(2^n)=n*2^(n-1)...