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 出即可
bn=2An-1+2^n-2=b(n-1)+1
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 出即可
bn=2An-1+2^n-2=b(n-1)+1