(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
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/29 21:03:10
(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
(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
(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
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^n
b[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*(1-1/2^n)/(1-1/2)=1+1-1/2^n=2-1/2^n
∴b[n]=2-1/2^(n-1)
2.a[n]/n=b[n]=2-1/2^(n-1)
∴a[n]=2n-n/2^(n-1)
s[n]=2(1+2+..+n)-(1/1+2/2+..+n/2^(n-1))
=n(n+1)-(1+1+3/2^2+..+n/2^(n-1))
设x=3/2^2+..+n/2^(n-1),则2x=3/2+..+n/2^(n-2)
∴x=2x-x=(3/2+..+n/2^(n-2))-(3/2^2+..+n/2^(n-1))
=3/2+1/2^2+..+1/2^(n-2)-n/2^(n-1)
=3/2+1/2^2*(1-1/2^(n-3))/(1-1/2)-n/2^(n-1)
=3/2+1/2-1/2^(n-2)-n/2^(n-1)=2-(n+2)/2^(n-2)
∴s[n]=n(n+1)-(2+x)=n(n+1)-4+(n+2)/2^(n-1)