sum of even integers

[sachindia]

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 ?

[Anurag@Gurome]

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 ?

There is a much better way to solve this without using the given formula.

For any set of uniformly spaced integers, the average of the integers = (Largest integer + Smallest integer)/2

In this case, all the even integers between 99 and 300, i.e. from 100 to 300 are uniformly spaced. Hence, average of them = (100 + 300)/2 = 200
Now, sum of all these integers = (Average)*(Number of integers in the range) = 200*[(300 - 100)/2 + 1] = 200*101 = 20,200

[Anurag@Gurome]

If anyone is interested in solving the problem using the formula...

Required sum = (100 + 102 + 104 + ... + 300) = 2(50 + 51 + ... + 150)

Now, (1 + 2 + ... + 150) = 150*151/2
And, (1 + 2 + ... + 49) = 49*50/2

Hence, (50 + 51 + ... + 150) = (150*151/2) - (49*50/2) = (50/2)*(3*151 - 49) = 50*404/2

Hence, required sum = 2*50*404/2 = 50*404 = 20,200