1. Note that a_k=C(n, k), so the general term of summation is k*C(n, k), so let's simplify it first:

2. A number divisible by 1 1 can be divisible by1if the difference satisfies the sum of odd and even numbers.

Divide eight numbers into two groups, each with four. Because the maximum difference between the two groups is (5+6+7+8)-(1+2+3+4) =16.

So the difference can be 1 1 or 0.

Let odd arrays sum to s? , even arrays sum to s? If s? -S？ = 1 1, and s? +S？ =36 when s? ，S？ Not an integer.

So the difference can't be 1 1, only 0. At this time, s? =S？ = 18

Since each number can only appear in one group, we only need to consider all possible combinations of one of the eight numbers.

Investigate 8, ***4 combinations:

8? 7? 2? 1

8? 6? 3? 1

8? 5? 4? 1

8? 6? 3? 2

For each combination, there are two choices: odd or even.

Put 4 in odd digits! So the even number of digits is 4! A method

So, with a * *, there are 4×2×4 ways to place numbers! ×4! =4608 species

That is, 4,608 8-digit numbers, without repetition, can be divisible by 1 1.