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Natural Numbers

Date: 01/08/98 at 01:50:42
From: moonyz
Subject: Natural numbers

Please name me two ways of finding the sum of "n" natural numbers.
1+2+3+4... + n = ?       n > 1

Date: 01/13/98 at 11:46:13
From: Doctor Bruce
Subject: Re: Natural numbers


Here is a simple way to find the sum of the first  n  natural numbers.
Write the numbers in a row two different ways: one row in ascending 
order from 1 to n, and one in descending order, from n to 1.  Write 
the two rows under each other and add down, like a regular addition 

          1   +   2   +   3   +   ...   +  n-1  +   n
       +  n   +  n-1  +  n-2  +   ...   +   2   +   1

         n+1  +  n+1  +  n+1  +   ...   +  n+1  +  n+1.

Now add the "answer" row  across. Since every number there is  
n+1, and there are n of them listed, we get the sum n*(n+1). 
This represents twice the the sum we want, because we wrote the 
numbers out twice. So the answer is

   1  +  2  +  3  +  ...  + n-1 +  n  =  n*(n+1)/2

Now, you asked for two ways to find the sum. Here is another way to do 
it, but I warn you this one is a little more complicated, because you 
have to reason differently for n being even or odd.  

This time, just write out one row of numbers from 1 to n  

   1   +   2   +   3   +   ...   +  n-1  +   n

Then fold the row in half, tucking the righthand end underneath. If n 
is even, the folded row will look like this:

                                   fold ->---------->--------\
        1     +    2      +   ...   +  (n/2)-1  +   n/2       v
     +  n     +   n-1     +   ...   +  (n/2)+2  + (n/2)+1.    |

The sum adding down is n+1 for each pair of numbers. There are  n/2  
such pairs. So, the sum is (n/2)*(n+1), which is the same answer as 

Now, if n is odd, there is a little wrinkle. We can line up the n at
the end of the row under the 1 at the beginning, as we just did. But
the row won't fold exactly in half - one term sticks out on the top.  
It will look like this:

  1   +   2    +  ...  +  [(n-1)/2]-1  +  [(n-1)/2]  +  [(n-1)/2]+1
+ n   +  n-1   +  ...  +  [(n-1)/2]+3  +  [(n-1)/2]+2

Adding down, each pair of numbers again gives n+1. But there are now 
only (n-1)/2  such pairs, and there is a leftover number, [(n-1)/2]+1.  
The total of all this is

     [(n-1)/2]*(n+1) + [(n-1)/2]+1  =  n*(n+1)/2

You should try to work out that last line, to make sure you believe 
that it comes out to the same answer!

-Doctor Bruce,  The Math Forum
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Associated Topics:
High School Sequences, Series

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