Sum the even numbers between 1000 and 2000 inclusive

[Pual Black]

this is just an arithmetic series but with a small difference. i will show that below

The attempt at a solution
the general arithmetic formula
$S_N=\sum_{n=1}^\infty n$
for my problem
$S_N=\sum_{n=1000}^{2000} n$
i have to rewrite it so i will just add the even numbers

$S_N=2\sum_{n=500}^{1000} n = 2\frac{n}{2}(first\,term+last\, term) =2 \frac{n}{2}(500+1000)$
now i have a problem
i think (n=number of terms) is 500 but if put this number i will get a wrong answer
if i put n=501 i will the right answer 751500

i have a similar problem
Sum the integers between 1 and 1000 inclusive
a=1
d=1
n=1000

$S_N=\sum_{n=1}^{1000} n =\frac{n}{2}(first\,term+last\, term)= \frac{1000}{2}(1+1000) =500500$

so why in problem one i have to put n=501 and in problem two i put n=1000
i think they are just the same but different values

i thought maybe n= last term - first term + 1
is that right?
sorry but i just started studying Infinite Series

 

[PietKuip]

Why use a formula? Just do the sum like this:
1000 + 2000 = 3000
1002 + 1998 = 3000
1004 + 1996 = 3000
...

 

[ArcanaNoir]

The number of terms from 500 to 1000 including both 500 and 1000 is actually 501. To make it easier to see, subtract 500 so we are counting the number of terms between 0 and 500 including both 0 and 500. Since you are starting at 0, and not 1, you actually have one extra term. In your second problem you start at 1 and go to 1000, so that works like you would expect and is just 1000 terms. If you started at 0 and went to 1000, you would have 1001 terms.

 

[andrewkirk]

Your method is OK but you have to be summing an even number of numbers. The number of even numbers between 1000 and 2000 inclusive is odd, so the method doesn't work. So instead sum over the range 1002 to 2000, and then add 1000 to the result.

 

[HallsofIvy]

You are missing the point. You want to sum even numbers from 1000 to 2000. So sum all integers from 500 to 1000 and multiply by 2.

 

[vela]

I thought maybe n= last term - first term + 1
Is that right?

Yes, that's right if d=1. If you're not sure, it's useful to check limiting cases to see if it's right. Consider a series with only one term so that first=last. You can see why the +1 has to be there.

 

[vela]

Your method is OK but you have to be summing an even number of numbers.

That's not true. The formula the OP is using works for both even and odd n.

 

[Pual Black]

Why use a formula? Just do the sum like this:
1000 + 2000 = 3000
1002 + 1998 = 3000
1004 + 1996 = 3000
...

This way i will spend hours just tipping numbers. Or i use your idea and make a formula from it.

The number of terms from 500 to 1000 including both 500 and 1000 is actually 501. To make it easier to see, subtract 500 so we are counting the number of terms between 0 and 500 including both 0 and 500. Since you are starting at 0, and not 1, you actually have one extra term. In your second problem you start at 1 and go to 1000, so that works like you would expect and is just 1000 terms. If you started at 0 and went to 1000, you would have 1001 terms.

Ok i will just make this in future
n= last term - first term + 1

Thank you all for your fast answers

 

[Pual Black]

Yes, that's right if d=1. If you're not sure, it's useful to check limiting cases to see if it's right. Consider a series with only one term so that first=last. You can see why the +1 has to be there.

Yes if i make this
n= last term - first term +1= 1-1+1 =1
And actually the number of terms is just 1

And if d is not equal to 1
I will make the same strategy and divide by d at the end

For example

First term = 1
Last term 1000
d=2
So n = 500