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项的和已知数列an满足a1=1,an=2a(n-1)+2^(n+1)+1,证明an+
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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