Natural NumbersDate: 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. 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 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/