在数列An中,A1=1,An+1=2An+2的n次方.（1）设Bn=An/2的（次方减1）,证明：Bn是等差数列.（2）求数列An

1,A(n+1)=2An+2^n,两边除以2^n得A(n+1)/2^n=An/2^(n-1)+1,即B(n+1)=Bn+1,Bn是等差数列.2,B1=A1=1,则Bn=n,即An=n2^(n-1)Sn=1+2*2^1+3*2^2+.+n2^(n-1)2Sn=2+2*2^2+3*2^3+.+n2^n,相减得Sn=n2^n-(1+2^1+2^2+...+2^(n-1...