winniethepooh wrote:For any positive integer n, the sum of first n positive integer equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?
A) 10,100
B) 20,200
C) 22,650
D) 40,200
E) 45,150
Again, since you want additional approaches (the ones Ashley suggests are perfectly good, and are most likely what I'd use), you can also estimate here easily enough. We're adding roughly half of the numbers from 100 to 300, so we're adding roughly 100 numbers with an average of 200, so the sum should be roughly 100*200 = 20,000. This estimate is only off because we left out one number, so B will be correct.
There are also many ways to rewrite the sum. We need to find:
100 + 102 + ... + 298 + 300 = 2(50 + 51 + ... + 149 + 150) = 2[(49 + 1) + (49 + 2) + (49 + 3) + ... + (49 + 100) + (49 + 101)]
Since we are adding the numbers from 1 through 101 in brackets, we have 101 forty-nines, so this is equal to
2[101*49 + 1 + 2 + 3 + ... + 100 + 101]
Now we can use the formula in the question with n=101:
= 2[101*49 + (101)(102)/2] = (2)(101)(49) + (101)(102) = (2)(101)[49 + 51] = 202*100 = 20,200
That's probably not the easiest approach, but you asked for "as many solutions as possible!"
