If I was to work out the sum of all the even numbers between 1000 and 2000, am I correct in saying that there are exactly 600 even numbers?
Therefore, is the final answer 959400?
Could someone please confirm this?
Sum = n/2[(2a+(n-1)d]
where n is the number of terms, a is the first term and d is the difference between each term.
The Attempt at a Solution