So the numbers of the triangle are 1, 3, 6, 10,1,28, 36, 45, 55, 66, 78, 9 1,/kloc-0.

So knowing that the number of triangles divisible by 5 appears twice in every five numbers, that is, every five numbers are divided into a group, then the last two numbers of the group can be divisible by 5.

Since b20 14 is a number divisible by 5 of 20 14,

Therefore, it appears in the last number of the 1007 group grouped by five segments in the series {an}.

Therefore, b20 14 is the number in the sequence {an} 1007×5=5035.

(2) Since 2k- 1 is an odd number, it can be known from (1) that the number 2k- 1 divisible by 5 appears in the penultimate position of the K group, so it is the term of k×5- 1=5k- 1 in the sequence {an}, so B2B. 1)2

So the answer is: 5035,5k (5k? 1)2