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Problem 9

Consider the sequence of sets with the following properties:

1. Each set contains only consecutive integers.
2. The nth set has one more element than the (n-1)th set.
3. The least element of the nth set is one more than the greatest element in the (n-1)th set.
4. The first set is {1}.

In other words, consider the sets {1}, {2,3}, {4,5,6}, {7,8,9,10}, ...

Now let S(n) be the sum of the elements in the nth set.

Find a formula in terms of n for S(n).

Solution

Let a(n) be the first element in the nth set. Notice that there are a(n)-1 elements in all the previous n-1 sets. So, a(n) is the sum of the number of elements in the previous plus one. So,

===.

Now, let b(n) be the last element in the nth set. Since there are n elements in the set, we know b(n)=a(n)+(n-1). Thus,

==.

So, the average element in the nth set is

and the sum of all the numbers in the set is n times the average element.

Therefore,

===.