We can solve this question even if we don't know any formula for such sums:

First even minus first odd = 2-1 = 1;
The sum of first 2 even integers minus the sum of first 2 odd integers = (2+4)-(1+3) = 2;
The sum of first 3 even integers minus the sum of first 3 odd integers = (2+4+6)-(1+3+5) = 3;
...
We can see the patterns here, so the sum of first 50 positive even integers minus the sum of first 50 positive odd integers will be 50.

Answer: C.

OR: each even minus its preceding odd is one, so x-y=50 (x-y=(even1+even2+...+even50)-(odd1+odd2+..+odd50)=(even1-odd1)+(even2-odd2)+...+(even50-odd50)=1+1+...+1=50).

Answer: C.

Hope it's clear.
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Sum of n first positive integers: 1+2+...+n=\frac{1+n}{2}*n. So, the sum of 100 first positive integers is (1+100)/2*100.

Sum of n first positive odd numbers: a_1+a_2+...+a_n=1+3+...+a_n=n^2, where a_n is the last, n_{th} term and given by: a_n=2n-1. Given n=5 first odd positive integers, then their sum equals to 1+3+5+7+9=5^2=25.

Sum of n first positive even numbers: a_1+a_2+...+a_n=2+4+...+a_n=n(n+1), where a_n is the last, n_{th} term and given by: a_n=2n. Given n=4 first positive even integers, then their sum equals to 2+4+6+8=4(4+1)=20.

For more check here: math-number-theory-88376.html

Hope it helps.
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Shortcut way without even a single calculation:-

Sum of first 'n' even integers is given by - n(n+1)
Sum of first 'n' odd integers is given by - n^2

x = n(n+1) = 50 x 51
y= n^2 = 50 x 50
x-y = 50 (51-50) = 50 (1) = 50
Answer C

Hope It helps
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Bunuel, if I wanted to find the sum of the even integers between 26 and 62, inclusive, then the formula above for even integers n(n+1) would not work, because this formula is only for the first n even positive integers meaning we would need to start at 2. Is this the correct way to think about it?

X = 2+4+6+8+.... +100
Y = 1+3+5+7+.... +99

X-Y= (2-1) + (4-3) + (6-5) + .... +(100-99) there are 50 terms
X-Y= 1 + 1 + 1 + .... +1 50 terms
=> X-Y=50
Answer C

I'll post two ways to solve this question : The formal way (in order for you guys to better understand the theory behind it) and the GMAT way (within the 2-minute scope).

Let's start.

1st method : The GMAT way

Bunuel actually proposed the easiest and fastest method to solve this question. That is you should look for patterns through examples.

First 2 even numbers : 2, 4 => summed up : (2+4) = 6
First 2 odd numbers : 1, 3 => summed up : (1+3) = 4

Their difference will be 2.

First 3 even numbers : 2, 4, 6 => summed up : (2+4+6) = 12
First 3 odd numbers : 1, 3, 5 => summed up : (1+3+5) = 9

Their difference will be 3.

And so forth. So eventually the difference between the sum of the first 50 even integers and the first 50 odd integers will be 50. Which is answer choice C.

2nd method : The formal way

To use this method you should be familiar and comfortable with :

- The general form of an even integer which is 2n ;
- The general form of an odd integer which is 2n+1 ;
- Counting the number of consecutive integers within a list which is given by the following formula : (Last number - First number) + 1 ;
- The sum operator and manipulating it.

Therefore the difference between the sum of the 50 first even integers and the 50 first odd integers is written as such :


You'll notice two things :

- I chose to index the sums from 0 to 49 since the first even integer is 0 and the first odd integer is 1 ;
- I inverted the difference since I've written Y-X instead of X-Y. This is due to the general form of the odd integer which if left as the original question stem suggests would leave me with a (-1) instead of 1.

If we develop the difference above we get :


Which, unsuprisingly, yields 50 which is answer choice C.

Note that you can combine two sums if and only if they have the same index range (0 to 49 in both cases).

Hope that helped.