问题 解答题
等差数列{an}中,a2=8,S6=66
(1)求数列{an}的通项公式an
(2)设bn=
2
(n+1)an
,Tn=b1+b2+b3+…+bn,求Tn
答案

(1)设等差数列{an}的公差为d,则有

a1+d=8
6a1+15d=66
        …(2分)

解得:a1=6,d=2,…(4分)

∴an=a1+d(n-1)=6+2(n-1)=2n+4             …(6分)

(2)bn=

2
(n+1)an
=
1
(n+1)(n+2)
=
1
n+1
-
1
n+2
        …(9分)

∴Tn=b1+b2+b3+…+bn=

1
2
-
1
3
+
1
3
-
1
4
+…+
1
n+1
-
1
n+2
=
1
2
-
1
n+2
=
n
2n+4
                                   …(12分)

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