Question

已知数列｛an｝的前n项和为Sn=n²+2n（1）求数列的通项公式an（2）Tn=1/a1a2+1/a2a3+1/a3a4+...+1/ana

Answer 1

(1)an=Sn-S(n-1)=2n+1
(2)裂（拆）项法
Tn=1/(3*5)+1/(5*7)+....+1/[an*a(n+1)]
=1/2{1/3-1/5)+(1/5-1/7)+1/an-1/a(n+1)}
=1/2[1/3-1/a(n+1)]

Answer 2

an=Sn-S(n-1)=2n+1
Tn=1/(3*5)+1/(5*7)+....+1/[an*a(n+1)]
=1/2{1/3-1/5)+(1/5-1/7)+1/an-1/a(n+1)}
=1/2[1/3-1/a(n+1)]

Answer 3

(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)]