Author 
Message 
ggk618 Poster
Post Number: 20
 Posted on Sunday, 21 August, 2011  03:43 pm:  
I'm getting better. Not freaking out when I see these problems now, thanks to your help. This has me stumped. Q: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301? !. So I know they are really looking for the sum of even integers between 100 and 300. So if we are looking for the number of even integers between 100 and 300, wouldn't that equal to 100 integers? 2. The constant formula is n(n=1)/2. I thought I was going down the right path, but my answer doesn't match one of the 5 possibilities, which are: a. 10,100 b. 20,200 c. 22,650 d. 40,200 e. 45,120 Can you guide me to the right direction? Thanks!!!! 
cypher Prolific poster
Post Number: 268
 Posted on Sunday, 21 August, 2011  04:38 pm:  
Here's a way to work out how many numbers you have. For example if you divide the numbers 9,12,15,18,21 by 3 you can see that the list has the same number of numbers as 3,4,5,6,7 and subtracting 2 you see this list has the same number of numbers as 1,2,3,4,5. That's five numbers. You could try that with 100, 102, 104..........300. Once you know how many number there are you could think about this 100+102+104..........296+298+300 + 300+298+296..........104+102+100. 
