SUM OF INTEGERS

[francoisph]

Hi,

please any ideas?

for any positive integers 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?

[Rahul@gurome]

The sequence here will be 100 + 102 + 104 +...
The sum is given by (n/2)[2a + (n-1)d], where a is the first term and d is the common difference
a = 100, d = 2
Number of even integers between 99 and 301 = (300 - 100)/2 + 1 = 101

Therefore, sum of first n positive integers= (101/2)[200 + 100*2] = 101*200 = 20,200