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

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SUMMARY

The sum of all even numbers between 1000 and 2000 is calculated using the formula Sum = n/2[(2a+(n-1)d]. The correct number of even integers in this range is 501, not 600, as the sequence includes both endpoints. The final answer for the sum of these even numbers is 1010500, not 959400. This conclusion is based on the arithmetic series formula and the correct identification of the first term (1000) and the last term (2000).

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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
 
Last edited:
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brandon26 said:

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 a1[/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!
 

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