问题
解答题
设数列{an} 前n项和Sn=
(1)求数列{an} 的通项公式an. (2)若a=3,Tn=a1a2-a2a3+a3a4-a4a5+…+(-1)n-1anan+1,求T100的值. |
答案
解(1)∵sn= n(an+1) 2 sn+1= (n+1)(an+1+1) 2
Sn+1-Sn得2an+1=(n+1)an+1-nan+1(12分)
即(n-1)an+1=nan-1③
∴nan+2=(n+1)an+1-1④(4分)
④-③得nan+2-(n-1)an+1=(n+1)an+1-nan
⇒n(an+2+an)=2nan+1
∴an+2-an+1=an+1-an=an-an-1═a2-a1(6分)
而n=1时S1=
=a1,a1+1 2
∴a1=1,又a2=a=a1+d
∴{an} 为等差数列,公式d=a-1
故an=a1+(n-1)d=(n-1)(a-1)+1;(8分)
(2)∵a=3
∴an=2(n-1)+1=2n-1(10分)
故T100=a1a2-a2a3+a100a101
=a2(a1-a3)+a4(a3-a5)++a100(a99-a101)
=-4(a2+a4++a100)
=-4
=-100(3+199)=-20200(13分)(a2+a100)×50 2