已知数列an满足an=2an-1+2^n-1(n>=2),a1=5,bn=(an-1)/2^n,(1)证明:bn为等差数列(2)求数列an的前n项和S
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![已知数列an满足an=2an-1+2^n-1(n>=2),a1=5,bn=(an-1)/2^n,(1)证明:bn为等差数列(2)求数列an的前n项和S](/uploads/image/z/1549930-58-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3an%3D2an-1%2B2%5En-1%28n%3E%3D2%29%2Ca1%3D5%2Cbn%3D%28an-1%29%2F2%5En%2C%EF%BC%881%EF%BC%89%E8%AF%81%E6%98%8E%3Abn%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CS)
已知数列an满足an=2an-1+2^n-1(n>=2),a1=5,bn=(an-1)/2^n,(1)证明:bn为等差数列(2)求数列an的前n项和S
已知数列an满足an=2an-1+2^n-1(n>=2),a1=5,bn=(an-1)/2^n,(1)证明:bn为等差数列(2)求数列an的前n项和S
已知数列an满足an=2an-1+2^n-1(n>=2),a1=5,bn=(an-1)/2^n,(1)证明:bn为等差数列(2)求数列an的前n项和S
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
用错位相减法求出即可
bn=2An-1+2^n-2=b(n-1)+1
bn-bn-1=1则bn为等差数列