basso25@ wrote:i cannot follow the answer explanation; please break down in simple terms for me. thank you!

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

There's a formula for this, but I'm not a big fan of memorizing tons of formulas.
Here's one approach.

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150-50+1=101)

To evaluate 2(50+51+52+...+149+150) I'll add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101x200 = [spoiler]20,200 = B[/spoiler]

Cheers,
Brent

basso25@ wrote:i cannot follow the answer explanation; please break down in simple terms for me. thank you!

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

Alternatively, if we want to evaluate 2(50+51+52+...+149+150) (see above), we can evaluate the sum 50+51+52+...+149+150, and then double it.

Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49)

Now we use the formula:
sum of 1 to 150 = 150(151)/2 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = [spoiler]20,200 = B[/spoiler]

Cheers,
Brent

Another way to solve this problem...

Number of even integers between 99 and 301 = (301 - 99)/2 = 202/2 = 101
Now, average of these even integers = (first integer + last integer)/2 = (100 + 300)/2 = 400/2 = 200

Hence, sum of the integers = 101*200 = 20,200

The correct answer is C.