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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

Hello,

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
problem.

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

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

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
before.

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
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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