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

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The discussion centers on calculating the sum of all even numbers between 1000 and 2000, with an initial claim of 600 even numbers. However, it is clarified that there are actually more than 600 even numbers in this range. The formula for the sum of an arithmetic series is provided, emphasizing the importance of correctly identifying the number of terms. The final answer of 959400 is questioned based on the initial assumption of the number of even integers. The conclusion highlights the need for accurate interpretation of the range to determine the correct sum.
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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
 
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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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