已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N*)设数列{an}的前n项和为SN,是否存在实数M,使得对于任意的n属于N*,恒有SN
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/08 09:50:43
![已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N*)设数列{an}的前n项和为SN,是否存在实数M,使得对于任意的n属于N*,恒有SN](/uploads/image/z/13309557-69-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D3%2F2%2Can%2B1%3D%282n%2B3%29%2F%284n%2B2%29an%28n%E5%B1%9E%E4%BA%8EN%2A%29%E8%AE%BE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASN%2C%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E5%AE%9E%E6%95%B0M%2C%E4%BD%BF%E5%BE%97%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E7%9A%84n%E5%B1%9E%E4%BA%8EN%2A%2C%E6%81%92%E6%9C%89SN)
已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N*)设数列{an}的前n项和为SN,是否存在实数M,使得对于任意的n属于N*,恒有SN
已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N*)
设数列{an}的前n项和为SN,是否存在实数M,使得对于任意的n属于N*,恒有SN
已知数列an满足a1=3/2,an+1=(2n+3)/(4n+2)an(n属于N*)设数列{an}的前n项和为SN,是否存在实数M,使得对于任意的n属于N*,恒有SN
a(n+1)=1/2 (2n+3)/(2n+1) an
=1/2*1/2 *(2n+3)/(2n+1) *(2n+1)/(2n-1) *a(n-1)
=.
=(1/2)^n *(2n+3)/(2n+1) *(2n+1)/(2n-1) *.* 5/3 *a1
=(1/2)^n *(2n+3)/3 *3/2
=(2n+3) *(1/2)^(n+1)
an=(2n+1)*(1/2)^n
Sn=3*1/2 +5*(1/2)^2+7*(1/2)^3+.+(2n+1)*(1/2)^n
1/2 Sn=3*(1/2)^2+5*(1/2)^3+.+(2n-1)*(1/2)^n+(2n+1)*(1/2)^(n+1)
1/2Sn=Sn-1/2Sn=3*1/2 -2((1/2)^2+(1/2)^3+.+(1/2)^n) -(2n+1)*(1/2)^(n+1)
=3/2-(2n+1)*(1/2)^(n+1) +1/2 *(1-(1/2)^(n-1)) /(1-1/2)
=3/2+1-(1/2)^(n-1)-(2n+1)*(1/2)^(n+1)
=5/2 -(1/2)^(n-1)-(2n+1)*(1/2)^(n+1)
所以Sn=5-2(1/2)^(n-1)-2(2n+1)*(1/2)^(n+1)
=5-4(1/2)^n-(2n+1)(1/2)^n
=5-(1/2)^n (4+2n+1)
=5-(2n+5)*(1/2)^n
显然Sn