已知数列{an}中,a1=3,前n项和Sn=12(n+1)(an+1)−1
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/08 19:26:34
已知数列{an}中,a1=3,前n项和S
(Ⅰ):证明:∵Sn=
1
2(n+1)(an+1)−1,∴Sn+1=
1
2(n+2)(an+1+1)−1
∴an+1=Sn+1−Sn=
1
2[(n+2)(an+1+1)−(n+1)(an+1)]
整理,得nan+1=(n+1)an-1①
∴(n+1)an+2=(n+2)an+1-1②
②-①得:(n+1)an+2-nan+1=(n+2)an+1-(n+1)an
即(n+1)an+2-2(n+1)an+1+(n+1)an=0∴an+2-2an+1+an=0,
即an+2-an+1=an+1-an∴数列{an}是等差数列
(II)∵a1=3,nan+1=(n+1)an-1,
∴a2=2a1-1=5∴a2-a1=2,
即等差数列{an}的公差为2,
∴an=a1+2(n-1)=2n+1,(n∈N*)
1
2(n+1)(an+1)−1,∴Sn+1=
1
2(n+2)(an+1+1)−1
∴an+1=Sn+1−Sn=
1
2[(n+2)(an+1+1)−(n+1)(an+1)]
整理,得nan+1=(n+1)an-1①
∴(n+1)an+2=(n+2)an+1-1②
②-①得:(n+1)an+2-nan+1=(n+2)an+1-(n+1)an
即(n+1)an+2-2(n+1)an+1+(n+1)an=0∴an+2-2an+1+an=0,
即an+2-an+1=an+1-an∴数列{an}是等差数列
(II)∵a1=3,nan+1=(n+1)an-1,
∴a2=2a1-1=5∴a2-a1=2,
即等差数列{an}的公差为2,
∴an=a1+2(n-1)=2n+1,(n∈N*)
已知数列{an}中,a1=3,前n项和Sn=12(n+1)(an+1)−1
已知数列{an}中,a1=1,前n项和Sn=n+23an
已知数列{an}中,a1=1,前n项和Sn=(n+2/3)an.
已知等差数列{an}中,a1=1,a3=-3.数列{an}的前n项和Sn.
已知数列{an}a1=2前n项和为Sn 且满足Sn Sn-1=3an 求数列{an}的通项公式an
已知数列{An}首项A1=2/3,An+1=2An/An+1,求数列{n/An}的前n项和Sn
已知数列(An)满足A1=1 An+1=3An 数列(Bn)前n项和Sn=n*n+2n+1
已知数列an首相a1=3,通项an和前n项和SN之间满足2an=Sn*Sn-1(n大于等于2)
已知数列{an}的前n项和为Sn,且a1=1,an+1=1/3Sn,
数列{an}的前n项和为Sn,已知A1=a,An+1=Sn+3^n(三的n次方),n∈N*
已知在数列an中,a1=1/2,an+1=3an/an+3,已知bn的前n项和为sn
设数列An的前n项和为Sn,已知a1=1,An+1=Sn+3n+1求证数列{An+3}是等比数列