问题描述: 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sn 1个回答 分类:综合 2014-10-30 问题解答: 我来补答 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 展开全文阅读