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Adding Arithmetic Sequences

From: Norma
Subject: Adding consecutive numbers

How do you add the numbers from 1 to 5000 without actually doing it or 
using a calculator?

I added 1 and 2 and got 3. 3 and 3 and got 6, but I know this takes 
way too long. Help!

From: Doctor Pat
Subject: Re: Adding consecutive numbers


If you wrote out all the numbers from 1 to 5000 and then wrote them 
backwards underneath, you would have twice as many numbers as you 
needed, but the problem is easier, here's why:

      1     2     3     4 ............... 4998  4999  5000
   5000  4999  4998  4997 ...............    3     2     1
   5001  5001  5001  5001 ............... 5001  5001  5001

Notice that if we add the two lists, we get a list that is the same 
number, 5001, repeating. In fact, since each of the lists is 5000 
numbers long, we have, in the sums, a list of 5000 numbers that are 
each 5001.  

You have to admit that adding 5000 5001's is a lot quicker than the 
other way since 5000 x 5001 = 25,005,000.

But wait, you say, that's too much. We were only supposed to add one 
list and we added two. Okay, then the answer must be half as much:
   1 + 2 + 3 + ... + 5000 = 25,005,000/2 = 12,502,500

See, you can do the whole thing in your head.  

You can use this method if you have any number of consecutive numbers, 
whether they start with 1 or not. In fact, the numbers do not have to 
even be consecutive. They just have to be in an arithmetic sequence, 
that is, the difference between any two adjacent numbers must 
always be the same (in your example the difference was 1).  

Let's add up all the odd numbers from 1 to 25. We do the same thing as 

     1   3   5  .........  21  23  25
    25  23  21  .........   5   3   1
    26  26  26  .......... 26  26  26  
We have to know how many numbers there are, and in this case there are 
13 twenty-sixes, so the total must be 13*26/2.  

In general if you have an arithmetic sequence of N numbers and you 
know the first and last one, you can find the sum by: 

   Sum = N(First + Last)/2   

Using this for your original set with N = 5000, first = 1 and 
last = 5000, we get the same result 5000(5001)/2.  

Hope that helps. Thanks for writing.
- Doctor Pat, The Math Forum
Check out our web site!   
Associated Topics:
High School Number Theory
High School Sequences, Series
Middle School Number Sense/About Numbers

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