Confirming Final Answer for Sum of Even Numbers Between 1000 and 2000 | 959400"

[brandon26]

Homework Statement

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?
Thank you.

Homework Equations

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

a=1000
d=2
n=600

 

[HallsofIvy]

Homework Statement

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?

Why would you think there are "exactly 600 even numbers" in 999 consective integers?

Therefore, is the final answer 959400?

Could someone please confirm this?
Thank you.

Homework Equations

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

a=1000
d=2
n=600

The answer depends upon whether "between 1000 and 2000" means "including 1000 and 2000" or not.

Another very nice formula for the sum of an arithmetic series is
$$n\left(\frac{a_1+ a_n}{2}\right)$$
where a_{1[/sup] and an are the first and last numbers in an arithmetic sequence of n numbers. However, there are a lot more than 600 even numbers between 1000 and 2000!}

 

[Architeuthis Dux]

Using the equation provided, we can calculate the sum of all even numbers between 1000 and 2000 as follows:

Sum = 600/2[(2*1000+(600-1)*2]
= 600/2[(2000+1198]
= 600/2[3198]
= 600/2*3198
= 600*1599
= 959400

Therefore, your final answer of 959400 is correct. There are indeed 600 even numbers between 1000 and 2000, and their sum is 959400. This can also be confirmed by manually counting the numbers or by using a calculator. Great job!