I NEED HEP PLEASE! I'M DESPERATE, 10 POINTS!
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I NEED HEP PLEASE! I'M DESPERATE, 10 POINTS!

[From: ] [author: ] [Date: 11-08-25] [Hit: ]
Find the sum of the consecutive integers.I would appreciate it so much if you showed the work to how you got the answer to the problem! Oh, and 10 points to the best answer!-Here is how Gauss did it.There are 75 integers from 1 to 75.......
What is the sum of these consecutive integers?

Find the sum of the consecutive integers.

From 1 to 75

From 1 to 833

From 11 to 60

From 50 to 90

From 198 to 299

From 4321 to 4421

I would appreciate it so much if you showed the work to how you got the answer to the problem! Oh, and 10 points to the best answer!

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Here is how Gauss did it.
There are 75 integers from 1 to 75.

Let x be the sum of the integers from 1 to 75.
1 + 2 + 3 + ... + 75 = x.

Reverse the series, and you get the same sum:
75 + 74 + 73 + ... + 1 = x

Add the equations together and you get 76 added 75 times:
http://www.flickr.com/photos/dwread/6077…

76×75 = 5700
But every number was used twice, so divide this sum by 2 to get 1 + 2 + ... + 75 = 2850.

The sum of integers from a to b = (b-a+1)(a+b)/2

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a) 2850
b) 347 361
c) 1 775
d) 2 870
e) 25 347
f) 441 471

Firstly, find the average number, so for the first example the average is (1+75)/2, which is 38. Then find out how many numbers there are, which is done by the Last Number – First Number + 1. So, (75-1+1) = 75. And finally multiply those two numbers together. 38*75 = 2850. The website I referenced might explain it better then I did. Hope this helped.

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The formula is n(n-1) all divided by 2
From 1 to 75 = (75)(76)/2 = 2850

Now finish the rest
1
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