Let's call the first odd integer "a". The rest of the odd integers can be expressed as a+2, a+4, a+6, … , a+2(n-1), for a total of n integers.

The arithmetic mean of these integers is equal to their sum divided by the number of integers, so we have:

y = (na + (2+4+6+…+2(n-1)))/n

Notice that 2+4+6+…+2(n-1) = 2(1+2+3+…+(n-1)) = 2(n-1)(n-1+1)/2 = n(n-1).

Substituting and multiplying both sides by n yields:

yn = na + n(n-1)

Dividing both sides by n, we have:

y = a+ n-1

The sum of the smallest and largest integers is a + a+2(n-1), or 2a+2(n-1)=2(a+n-1)=2y.

Hence the answer is 2y.
### Instruction:
The arithmetic mean of an odd number of consecutive odd integers is y. Find the sum of the smallest and largest of the integers in terms of y.

### Response:
