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What is the sum of positive integers from 1 to 1000?

## What is the sum of positive integers from 1 to 1000?

[From: Mathematics] [author: ] [Date: 06-15] [Hit: ]
What is the sum of positive integers from 1 to 1000?What is the sum of positive integers from 1 to 1000? How would I figure this out step by step?......

What is the sum of positive integers from 1 to 1000?
What is the sum of positive integers from 1 to 1000?

How would I figure this out step by step?
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Puzzling say: As others have noted, if the first number is 1 and the last number is 1000, the average is 500.5. All the other numbers will pair the same way (2 and 999, 3 and 998, etc.) with the same average.

If you have 1000 numbers that average to 500.5, the sum must be 500,500.

In general, the sum of the first n positive integers is:
n(n + 1)/2

For n=1000:
1000 * 1001 / 2
= 1000 * 500.5
= 500,500

And if you are only dealing with powers of 10, another shortcut is to take half the number and repeat it.
10 --> 55
100 --> 5050
1000 --> 500500
10000 --> 50005000
100000 --> 5000050000
etc.

500,500
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Jim Moor say: Add the same numbers twice, the second row in reverse order

s = 1 + 2 + 3 + 4 + .... + 997 + 998 + 999 + 1000
s = 1000 + 999 + 998 + 997 + .... + 4 + 3 + 2 + 1
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2s = 1001 + 1001 + 1001 etc 1000 times
2s = 1001 * 1000

2s = 1001000
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JSG say: 1,000*1001/2 = 500500
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John say: If you add them up one at a time you might find out something interesting and get the correct answer at the same time.
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busterwasmycat say: the simple way is to define the midpoint and generate pairs that make the same total. 500+501 is the same quantity as 1000+1. you would have 500 such pairs. so, the total must be 500x1001.

The only real trick is that some ranges have an odd number of items and the midpoint value has no equivalent (cannot be paired) so you have to add it separately. like in add 1 through 999; 500 is the median value. you would have 499x1000+500 as a total. or, from the other way, 500x1000-500. (double the median value to make an extra pair, and then subtract it, or its half-pair value)
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Φ² = Φ+1 say: The sum from 1 to 1 is 1 , which is 0.5 + 0.5
The sum from 1 to 10 is 1 + 2 + 3 + ... + 10 = 55
The sum from 1 to 100 is 1 + 2 + 3 + ... + 100 = 5050
Based on the pattern, the sum from 1 to 1000 is probably 500500
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Ray say: One way is to treat it as an Arithmetic Series for which a = 1, d = 1 and n= 1000.

Sum = (n/2)[2a + (n - 1)d]

Sum = 500(2 + 99 x 1)

Sum = 500 x 101

Sum = 50500
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Then it's 1 + 999 = 1000
and .. 2 + 998 = 1000
3 + 997 = 1000
4 + 996 = 1000
etc etc until you get to 499 + 501
leaving yiou with the 500
It will take 500 steps to get here
so multiply 1000 by 500
and add the 500 which will be 'on it's own, in the middle'.

500x1000 + 500
500500

@ the comment by amania_r .. I DID ADD THE INITIAL 1000
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Bone Alone say: 1 + 1000 = 1001
2 + 999 = 1001
3 + 998 = 1001

See a pattern developing.

By the time you get to 500 + 501, you would have used up all the numbers from 1 to 1000.

So 500(1001) gives you the answer.
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Quentin say: There is a trick to this.
They can be laid out as a triangle.
*
**
***
Which is half a rectangle
****
****
****
The above is for 1 to 3. The sum is 3*4/2

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Puzzling say: As others have noted, if the first number is 1 and the last number is 1000, the average is 500.5. All the other numbers will pair the same way (2 and 999, 3 and 998, etc.) with the same average.

If you have 1000 numbers that average to 500.5, the sum must be 500,500.

In general, the sum of the first n positive integers is:
n(n + 1)/2

For n=1000:
1000 * 1001 / 2
= 1000 * 500.5
= 500,500

And if you are only dealing with powers of 10, another shortcut is to take half the number and repeat it.
10 --> 55
100 --> 5050
1000 --> 500500
10000 --> 50005000
100000 --> 5000050000
etc.

500,500
-
Mike G say: 1001 * 500 = 500,500
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la console say: s = 1 + 2 + 3 + 4 + .... + 997 + 998 + 999 + 1000
s = 1000 + 999 + 998 + 997 + .... + 4 + 3 + 2 + 1

2s = 1001 * 1000
s = 1001 * 500
s = 1001 * 100 * 5
s = 100100 * 5
s = 500500
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SumDude say: FORMULA : last number (n) .... n*(n+1)/2 ... for any number (n) you pick. Variations ... say you want to do 2+4+6+8+10 first divide by 2 then use formula, then multiple answer by 2. {or, simplified, the *2 and /2 cancel out and 5*6 = 30)
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