# Link

http://gre.magoosh.com/questions/213

# Question

Can we use the sum of sequences formula?

# Answer

While in a way we can use the same strategy as shown in the sum of sequences video, we can't simply use the formula of (n(n+1))/2, because neither of these sequences are of consecutive integers 1,2,3,4.... That formula is limited to sequences of integers starting from 1 and increasing by 1.

In this case, since we're looking at *even* and *odd* integers, we can't just apply the formula. But there's an FAQ under that lesson video which explains a strategy. We'll copy-paste that here for easy reference:

Q. How can I use the formula for a sum of n consecutive integers (found at 2:39) to find the sum of numbers in a non-consecutive series, such as {2,4,6,8...100} or {3,5,7,9...101}?

*A. Using the formulas in this lesson for solving problems involving the sum of non-consecutive series may be a more efficient strategy than memorizing formulas for specific series, such as the even and odd series.*

*Looking at the even series 2 + 4 + 6 + 8 + ... + 100, we can break this down into 2 x (1 + 2 + 3 + 4 + ... + 50). In other words, we can factor out a 2 from each term. Using the sum of n consecutive integers formula, we can calculate the sum of 1 through 50: 50 x 51/2 = 25 x 51 = 1275. And we multiply that by 2 (the number we originally factored out). 1275 x 2 = 2550. This approach will work for all even numbered series.*

*For the odd series, we’ll take a slightly different approach since we can't just factor out a number. We know how to calculate the sum of the numbers from 1 to 100. And we also know how to calculate the sum of the even numbers from 1 to 100. So we can do the following: Sum of odd consecutive integers from 1 to 100 = (Sum of all consecutive integers from 1 to 100) - (Sum of even consecutive integers from 1 to 100).*

*Sum of odds = (100 x 101/2) - [2 x (50 x 51/2)] = 5050 - 2550 = 2500. Now, the set {3,5,7,9...101} requires us to consider 3 to 101 not 1 - 100. So we subtract 1 and add 101 to 2500 and get 2600.*

* *

So in this question, we would find that the sum of the first 30 positive even integers would be 2 * (1 + 2 + 3 + 4 + ... + 30). Using the formula, we can work that out:

2 * 30(30 + 1)/2

30(30 + 1)

30(31)

930

For the odd numbers, we need to see what the total list would be if both odds and evens were included, all the way up to the 31st odd number. We find that the 31st odd number must be 61. So using the sum of sequences formula, here's what we get:

61(61 + 1)/2

61(62)/2

61(31)

1,891

Subtract out the even sum, 930, and we're left with an odd sum of 1,891 - 930 = 961.

Finally, 961 - 930 = 31.

And here's another approach to the same idea:

Let's start with X, (1+3+5+...+57+59+61). If we add the first and last numbers in the series, we get 1+61 = 62. Similarly, adding the second and second-to-last numbers gives us 3+59 = 62. Next, 5+57 = 62. As we move inward, adding each pair of numbers, we get 62 every time. We know there are 31 numbers in this set, so we can use 15 pairs of numbers. 15*62 = 930.

But in the middle of that set, there's one more number without a partner, 31. When using this strategy, if there's a number left in the middle, it's always half the sum of each pair. That sum was 62, so the number in the middle is 31. So now we have 930 + 31 = 961.

For the even numbers, Y, we follow the same process using (2+4+...+58+60), and we get (60+2) * 15 = 930.

And that means we have our answer:

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