在数列an中,a1=1.an+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Sn在数列an中,a1=1,an+1=(1+1/n)an +(n+1)/2^n (1)设bn=an/n,求数列{bn}的通项公式
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/12 08:31:28
![在数列an中,a1=1.an+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Sn在数列an中,a1=1,an+1=(1+1/n)an +(n+1)/2^n (1)设bn=an/n,求数列{bn}的通项公式](/uploads/image/z/1803972-12-2.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97an%E4%B8%AD%2Ca1%3D1.an%2B1%3D%EF%BC%881%2B1%2Fn%29an+%2B%EF%BC%88n%2B1%29%2F2%5En+%281%29%E8%AE%BEbn%3Dan%2Fn%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E5%9C%A8%E6%95%B0%E5%88%97an%E4%B8%AD%EF%BC%8Ca1%3D1%EF%BC%8Can%2B1%3D%EF%BC%881%2B1%2Fn%29an+%2B%EF%BC%88n%2B1%29%2F2%5En+%281%29%E8%AE%BEbn%3Dan%2Fn%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
在数列an中,a1=1.an+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Sn在数列an中,a1=1,an+1=(1+1/n)an +(n+1)/2^n (1)设bn=an/n,求数列{bn}的通项公式
在数列an中,a1=1.an+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Sn
在数列an中,a1=1,an+1=(1+1/n)an +(n+1)/2^n (1)设bn=an/n,求数列{bn}的通项公式
在数列an中,a1=1.an+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Sn在数列an中,a1=1,an+1=(1+1/n)an +(n+1)/2^n (1)设bn=an/n,求数列{bn}的通项公式
(1)
a(n+1)=(1+ 1/n)an+(n+1)/2ⁿ=[(n+1)/n]an+(n+1)/2ⁿ
a(n+1)/(n+1)=an/n +1/2ⁿ
a(n+1)/(n+1)-an/n=1/2ⁿ
an/n -a(n-1)/(n-1)=1/2^(n-1)
a(n-1)/(n-1)-a(n-2)/(n-2)=1/2^(n-2)
…………
a2/2-a1/1=1/2
累加
an/n -a1/1=1/2+1/2²+...+1/2^(n-1)=(1/2)[1-(1/2)^(n-1)]/(1-1/2)=1-1/2^(n-1)
an/n=a1+1-1/2^(n-1)=2- 1/2^(n-1)
bn=an/n bn=2-1/2^(n-1)
数列{bn}的通项公式为bn=2- 1/2^(n-1)
(2)
an=2n -n/2^(n-1)
Sn=a1+a2+...+an=2(1+2+...+n) -[1/1+2/2+3/2²+...+n/2^(n-1)]
令Cn=1/1+2/2+3/2²+...+n/2^(n-1)
则Cn/2=1/2+2/2²+...+(n-1)/2^(n-1)+n/2ⁿ
Cn-Cn/2=Cn/2=1+1/2+1/2²+...+1/2^(n-1)-n/2ⁿ
=1×(1-1/2ⁿ)/(1-1/2)-n/2ⁿ
=2-(n+2)/2ⁿ
Cn=4 -(n+2)/2^(n-1)
Sn=2(1+2+...+n) -Cn
=2n(n+1)/2 -4+(n+2)/2^(n-1)
=(n+2)/2^(n-1) +n²+n -4