已知数列an满足 a1=1,an=2a(n-1)+2^(n+1)+1,证明an+1/2^n为等差数列,并求出该数列前n项的和
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![已知数列an满足 a1=1,an=2a(n-1)+2^(n+1)+1,证明an+1/2^n为等差数列,并求出该数列前n项的和](/uploads/image/z/7282632-48-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3+a1%3D1%2Can%3D2a%28n-1%29%2B2%5E%28n%2B1%29%2B1%2C%E8%AF%81%E6%98%8Ean%2B1%2F2%5En%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82%E5%87%BA%E8%AF%A5%E6%95%B0%E5%88%97%E5%89%8Dn%E9%A1%B9%E7%9A%84%E5%92%8C)
已知数列an满足 a1=1,an=2a(n-1)+2^(n+1)+1,证明an+1/2^n为等差数列,并求出该数列前n项的和
已知数列an满足 a1=1,an=2a(n-1)+2^(n+1)+1,证明an+1/2^n为等差数列,并求出该数列前n项的和
已知数列an满足 a1=1,an=2a(n-1)+2^(n+1)+1,证明an+1/2^n为等差数列,并求出该数列前n项的和
an=2a(n-1)+2^(n+1)+1
an+1=[2a(n-1)+2]+2^(n+1) 【两边同除以2^n】
[an+1]/(2^n)=[a(n-1)+1]/[2^(n-1)]+2
即:[an+1)/[2^n]-[a(n-1)+1]/[2^(n-1)]=2=常数,所以数列{[an+1]/[2^n]}是以(a1+1)/2=1为首项,以d=2为公差的等差数列,则:
[an+1]/[2^n]=2n-1
an=[(2n-1)×2^n]-1
求an的前n项和,方法:1、先分组,2、第一组求和采用错位法求和.
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