(1)∵a2=1+d,a5=1+4d,a14=1+13d
∴(1+4d)2=(1+d)(1+13d)
∵d>0
∴d=2
∴an=1+2(n-1)=2n-1;
(2)bn=
=1 n(an+3)
(1 2
-1 n
),1 n+1
∴Sn=
(1-1 2
+1 2
-1 2
+…+1 3
-1 n
)=1 n+1
(1-1 2
)=1 n+1
;n 2n+2
(3)Sn>
,即t 36
>n 2n+2
,t 36
∴
≥1 2
,t 36
∴t≤18,
∴最大的整数t为18.