For any positive integer n,the sum of first n positive integers
equals n(n+1)/2 . What s the sum of all even integers between 99 and 301
equals n(n+1)/2 . What s the sum of all even integers between 99 and 301
We want 100+102+104+....298+300avinashsidha wrote:For any positive integer n,the sum of first n positive integers
equals n(n+1)/2 . What s the sum of all even integers between 99 and 301
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 =20,200
Cheers,
Brent
