1已知数列an满足an=2an-1+2^n-1(n>=2),有an-1=2（an-1-1）+2^n,两边同时除以2^n,得bn=bn-1+1故数列{bn}为首项b1=2,d=1的等差数列2由一问可知,an=（n+1)2^n+1故sn=n*（n+1)/2 +2*2+3*2^2+……+（n+1)*2^n用错位相减法...