For any positive integer...

Blues wrote:Hello,

Problems like this one from OG12 confuse me. I look at the solution provided, but I can't understand what the problem is asking me to do. I can mimic the work, but I doubt I'd understand how to apply it to another problem. Any help understanding the underlying concepts would be greatly appreciated:

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?

(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

Answer: B

This looks so easy, I'm a little embarrassed even asking, but I won't get any better at this hiding my weaknesses. Thanks for helping me on my quest for 700+

Dude you're fine.

you might want to memorize that the sum of consecutive evens is n(n+1) where n is number of terms.

for this particular problem

sum of evens=average of evens x number of even terms

given that we are looking for evens from 99 to 301. translation is evens from 100 to 300.

average of evens = (first number + last number)/2
=(100+300)/2=200

number of even terms = (last number - first number)/2 + 1 because its inclusive so
(300-100)/2 + 1
=101

sum of evens(99 to 301)=200 x 101=20200

OA is B

you got this!