数列an中,a1=1,an+1=2an+2的n次方,设bn=an／2∧n-1,证明bn是等差数列,求数列an的前n项和s

a(n+1)=2an+2^na(n+1)/2^n=2an/2^n+1a(n+1)/2^n=an/2^(n-1)+1a(n+1)/2^n-an/2^(n-1)=1,为定值。a1/2^(1-1)=1/1=1数列{an/2^(n-1)}是以1为首项,1为公差的等差数列。bn=an/2^(n-1)数列{bn}是以1为首项,1为公差的等差数列。an/2^(n-1)=1+(n-1)=nan=n×2^(n-1)Sn=a1+a2+。。。+an=1×2^0+2×2^1+。。。+n×2^(n-1)2Sn=1×2^1+2×2^2+。。。+(n-1)×2^(n-1)+n×2^nSn-2Sn=-Sn=2^0+2^1+2^2+。。。+2^(n-1)-n×2^n=(2^n-1)/(2-1)-n×2^n=2^n-1-n×2^n=(1-n)×2^n-1Sn=(n-1)×2^n+1