SOLUTION: The odd integers are arranged into sets, each succeeding set continuing from the previous, and containing one more number as follows: {1},{3,5},{7,9,11},{13,15,17,19},... what is t

Question 1099657: The odd integers are arranged into sets, each succeeding set continuing from the previous, and containing one more number as follows: {1},{3,5},{7,9,11},{13,15,17,19},... what is the sum of the numbers in the set that has last number 1259?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The numbers of numbers in the successive sets are
1, 2, 3, 4, ...

Let's think about the process for finding the answer by answering a similar but much simpler problem: find the sum of the numbers in the set that has last number 19.

We can see the answer in the statement of the problem; however, let's see how we can get the answer in a way that we can use to find the answer to the given problem.

The "first" set contains 1 odd integer; the "second" contains 2; the third 3; and the fourth 4.

1+2+3+4 = 10, and 19 is the 10th odd integer, so 19 is the last number in the 4th set.

Then, since there are 4 numbers in that set, the four numbers in the set are 19, 17, 15, and 13.

It is easy to see in this case that the sum of those numbers is 64. But let's find that sum by a method that we can use for the given problem, where we can't see the whole set of numbers at once.

We know at this point that 19 is the largest number in the 4th set. The smallest number in the 4th set is 19 minus the common difference 2 three times, so the smallest number in the set is . Then the average of the numbers in the set is ; and with 4 numbers in the set, the sum is .

Now let's apply that same analysis to the given problem.

The number 1259 is the 630th odd positive integer; and we are told that it is the last number in its set.

So we need to find when 1+2+3+4+...+n is equal to 630.

The sum of the integers from 1 to n is ; so we want to find a solution to

By one method or another, we find n=35.

That means 1259 is the last number in the 35th set.

Using the method we developed earlier, the first number in the 35th set is .

Then the average of the numbers in the set is .

And then the sum of the numbers in that set is .

ANSWER: The sum of the numbers in the set with last number 1259 is 42,875.

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