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 wayBunuel 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 wayTo 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 :
Attachment:
pic3.jpg [ 6.93 KiB | Viewed 43837 times ]
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 :
Attachment:
pic4.jpg [ 10.55 KiB | Viewed 43763 times ]
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.