Question

已知数列{an}的前n项和Sn=n2+2n(|)求数列的通项公式an(2)设Tn=a1a2/ |+a2a3/1+a3a4/1+…+anan+|/1,求Tn

(1)Sn=n2+2n
an=Sn-S(n-1)
=n2+2n-[(n-1)2+2(n-1)]
=2n+1
(2)1/ana(n+1)
=1/(2n+1)(2n+3)
=[1/(2n+1)-1/(2n+3)]/2
∴Tn=[1/3-1/5+1/5-1/7+1/7-1/9+……+1/(2n+1)-1/(2n+3)]/2
=[1/3-1/(2n+3)]/2
=n/[3(2n+3)]

为什么由1/ana(n+1)
得出1/(2n+1)(2n+3)

Answer 1

an=2n+1
a(n+1)=2(n+1)+1=2n+3
所以1/[ana(n+1)]=1/[(2n+1)(2n+3)]