(2009 全国)在数列{an}中,a1=1,a(n+1)=(1+1/n)an+(n+1)/2^n.1.设bn=an/n,求数列｛_数学解答

(2009 全国)在数列{an}中,a1=1,a(n+1)=(1+1/n)an+(n+1)/2^n.
1.设bn=an/n,求数列｛bn｝的通项公式
2.求数列{an}的前n项和s

人气：486 ℃　时间：2020-09-21 11:10:08

解答

1.a[n+1]=((n+1)/n)a[n]+(n+1)/2^n所以:a[n+1]/(n+1)=a[n]/n+1/2^n即:b[n+1]=b[n]+1/2^nb[n]=b[n-1]+1/2^(n-1),..,b[2]=b[1]+1/2将以上n式相加得到:b[n+1]+b[n]+..+b[2]=b[n]+..+b[1]+(1/2^n+..+1/2)b[n+1]=b[1]+1/2...