Expert: Paul Klarreich - 2/3/2006

Question
Hello,

I'm working on a proof for mathematical induction that the sum of the first n numbers is equal to n(n+1)/2.  What is meant in this question by the “…sum of the first n numbers…”.  I'm interpretting it as (n^2 + n) = n(n+1)/2, but this only works for 0,1 and after that it's an inequality.  When I used incrementally 0,1,2,3,4,5, and 6 I obtained 0,1,3,6,10,15, and 21 respectively, but this doesn't seem to have show any special pattern to me.  I can definitely see how n(n+1)/2 is a constant, but I'm confused about what is meant by  “…sum of the first n numbers…” in this particular instance.  What's a simple example I could start with?

Eddie.  

Answer
Hi, Eddie,

Your Question:  Hello,

I'm working on a proof for mathematical induction that the sum of the first n numbers is equal to n(n+1)/2. What is meant in this question by the “…sum of the first n numbers…”.
WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

>> That means you are to prove that:
                     n(n+1)
1 + 2 + 3 + ... + n = ------
                       2


I'm interpreting it as (n^2 + n) = n(n+1)/2, but this only works for 0,1 and after that it's an inequality.

>> Of course it is.  n(n+1)/2 = (n^2 + n)/2, which is not the same as n^2 + n


When I used incrementally 0,1,2,3,4,5, and 6 I obtained 0,1,3,6,10,15, and 21 respectively, but this doesn't seem to have show any special pattern to me.

>> It should.  These are all what are called 'triangular numbers'.  Did you ever go bowling?  The tenpins, excuse me, ten pins are arranged in a triangular fashion with four pins on a side.  This is the fourth triangular number (counting from one, that is, not from zero) and is 4(5)/2.  

I can definitely see how n(n+1)/2 is a constant, but I'm confused about what is meant by “…sum of the first n numbers…” in this particular instance. What's a simple example I could start with?
------------------------------------------
This one is simple enough.  The sum of the first n INTEGERS would be written as

1 + 2 + 3 + ... + n

This means to write a sum of integers where the first is 1 and the last is n, and you write only as many as are needed.

n = 7?  Write 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28
n = 3?  Write 1 + 2 + 3                 = 6
n = 1?  Write 1                         = 1
n = 0?  Don't write anything, and the sum is zero.

Now then, do you need the proof that

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

or can you finish it yourself?  If not, and MI is troublesome for you (it is, for many people) then let me know and I'll send it along.