在数列｝an｝中,a1=2,an=2an-1+2^n+1（n》=2） 令bn=an/2^n,求证｛bn｝是等差数列.

b（n+1）=a（n+1）/[2^(n+1)]=[2an+2^(n+2)]/[2^(n+1)]=[an+2^(n+1)]/(2^n)
b(n+1)-bn=[an+2^(n+1)]/(2^n)-an/(2^n)=[2^(n+1)]/(2^n)=2
所以{bn}是等差数列，首项b1=a1/2=1，公差为2

an/2^n
=(2an-1)/2^n + 1
=(an-1)/2^(n-1)+1
an/2^n - (an-1)/2^(n-1) = 1
则an/2^n是公差为1的等差数列

bn-bn-1=an/2^n-an-1/2^(n-1)
=(2an-1+2^n+1)/2^n-2an-1/2^n
=(2^n+1)/2^n
（2^n+1)/2^n不是一个常数，所以Bn不是等差数列

你只需要将an用bn替换然后代入左边an=2an-1+2^n+1再同时除2的n次方就可以了，得公差为2

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...