设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数列...设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 20:53:06
![设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数列...设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数](/uploads/image/z/5090268-12-8.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%943Sn%3Dan%2B4.%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3bn%3D3Sn%2C%E6%B1%82%E6%95%B0%E5%88%97...%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%943Sn%3Dan%2B4.%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3bn%3D3Sn%2C%E6%B1%82%E6%95%B0)
设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数列...设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数
设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数列...
设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数列{bn}的前n项和Tn
设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数列...设数列{an}的前n项和为Sn,且3Sn=an+4.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=3Sn,求数
【一】
当n=1时,3S1=a1+4,因S1=a1,则:a1=2;
当n≥2时,由:3Sn=an+4及3S(n-1)=a(n-1)+4
两式相减,得:
3an=an-a(n-1)
[an]/[a(n-1)]=-1/2=常数
则数列{an}是以a1=2为首项、以q=-1/2为公比的等比数列,得:an=2×(-1/2)^(n-1)
【二】
3Sn=an+4,且bn=3Sn,则:
bn=an+4
Tn=(a1+a2+…+an)+(4+4+…+4)=Sn+4n=(1/3)(an+4)+4n=(1/3)[2×(-1/2)^(n-1)+4]+4
(1). n=1时;
3a1=a1+4 —> a1=2
n>=2时;
3Sn=an+4 (1)
3S(n-1)=a(n-1)+4 (2)
(1)-(2)得 3an=an-a(n-1) —> an=-1/2 *a(n-1) = (-1/2)^(n-1)...
全部展开
(1). n=1时;
3a1=a1+4 —> a1=2
n>=2时;
3Sn=an+4 (1)
3S(n-1)=a(n-1)+4 (2)
(1)-(2)得 3an=an-a(n-1) —> an=-1/2 *a(n-1) = (-1/2)^(n-1) *a1
所以数列{an}的通项为:an =(-1)^(n-1)(1/2)^(n-2)
(2). 由(1)得 {an}是以 q=-1/2 的等比数列
Sn=a1+a2+...+an=4-(1/2)^(n-2)
Tn=b1+b2+...+bn=3*[S1+S2+...+Sn]=3*{4n-4*[1-(1/2)^(n-1)]}
——> Tn=12(n-1)+12(1/2)^(n-1)
收起