We give three proofs here that the nth Triangular number, 1+2+3+...+n is n(n+1)/2.The first is a visual one involving only the formula for the area of arectangle. This is followed by two proofs using algebra. The first uses "..." notation and the second introduces you to theSigma notation which makes the proof more precise.
A visual proof that 1+2+3+...+n = n(n+1)/2
We can visualize the sum 1+2+3+...+n as a
triangle of dots. Numbers which have such a pattern of dots are called
Triangle (or triangular) numbers, written T(n), the sum of the integers from 1 to n :
n  1  2  3  4  5  6 
T(n) as a sum  1  1+2  1+2+3  1+2+3+4  1..5  1..6 

T(n) as a triangle      ...  

T(n)=  1  3  6  10  15  21 

For the proof, we will
count the number of dots in T(n) but, instead of
summing the numbers 1, 2, 3, etc up to nwe will find the total using only
one multiplication and one division!To do this, we will fit two copies of a triangle of dots together, one red and an upsidedown copy in green.
E.g. T(4)=1+2+3+4
Notice that  we get a rectangle which is has the same number of rows (4) but has one extra column (5)
 so the rectangle is 4 by 5
 it therefore contains 4x5=20 balls
 but we took two copies of T(4) to get this
 so we must have 20/2 = 10 balls in T(4), which we can easily check.
This visual proof applies to any size of triangle number.
Here it is again on T(5):
So T(5) is half of a rectangle of dots 5 tall and 6 wide, i.e. half of 30 dots, so T(5)=15.
Try the formula for yourself with this Quiz (click on the button) which opens in a new window. After the Quiz, close its window and try this button again for another Quiz question.
For T(n)=1+2+3+...+n we take two copies and get a rectangle that is n by (n+1).
So there you have it  our visual proof that
T(n)=1 + 2 + 3 + ... + n=n(n + 1)/2
The same proof using algebra!
Here's how a mathematician might write out the above proof using algebra:
T(n)+T(n)  =  1 +  2 +  3 +  ...+  (n1) +  n 
 +  n +  (n1) +  (n2) +  ... +  2 +  1  Two copies, one red and the other, reversed, in green 
 =  (1 + n) +  (2 + n1) +  (3 + n2) +  ... +  (n1 + 2) +  (n + 1)  pair off the terms, a red with a green 
 =  (n+1) +  (n+1) +  (n+1) +  ... +  (n+1) +  (n+1)  All the n pairsums are equal to (n+1) 
2 T(n)  =  n (n+1) 
T(n)  =  n (n+1) / 2 
Using the Sigma notation
Some people regard the "..." as too vague and want a more precise alternative. For this reason, in summing a series, the
sigma notation is used.
Sigma is the name of the greek letter for the English "s", written as
(like an M on its side)as a capital letterand
(like a small b that's fallen over) in lower case. In this case, the "s" stands for "sum". (A tall curly form of S gives themathematical symbol for
integration  another kind of sum).
Mathematicians use the capital sigma for the sum of a series as follows: a formula describes the i^{th} term of the series being summed. It is written after the sigma;
 the starting value for i is written below the sigma;
 the ending value for i is written above the sigma
  i=final value  (formula for i^{th} term)   i=starting value 

In fact, the formula after the sigma can be written in terms of any variable not just i, for instance k, but then we must indicate whichis the letter that varies in the sum under the sigma.
Often the variable is omitted above the sigma but never omitted below the sigma.
Here are some examples:
The sum 10^{2}+11^{2}+12^{2}, where the numbers added are the square numbers i^{2}:  i=12  i^{2}   i=10 

The same sum can also be written in many other ways, for instance,as the sum of the square numbers (i+9)^{2}where this time i goes from 1 to 3  i=3  (i+9)^{2}   i=1 

or as the sum of the square numbers (i+11)^{2} where this time i goes from 1 to 1 (i.e. i = 1, 0 and 1)  i=1  (i+11)^{2}   i=1 

The sum 1+2+3+..+9 is T(9) or   i=9  i  T(9) =    i=1 

Here is T(n) which is 1+2+3+...+n, this time omitting the second use of the i above the sigma:  n  i = T(n)   i=1 

and this time, we have T(n) but written backwards: n+ (n1) + ... 3 + 2 + 1 where the i^{th} term is now n+1i for i from 1 to n:  i=n  (n+1i)=T(n)   i=1 

Finally, note that if all the terms are independent of the variable, for instanceif there is no i in the formula but the variable below the sigma is i, then all the terms are constant. The number of terms will begiven by the starting and ending values. Here, all the terms are fixed (constant) at 3:  i=7  3 = 3+3+3+3 = 12   i=4 

Here is the algebraic proof from above but now written using the sigma notation:
T(n)+T(n)  =  i=n  i   i=1 
 +  i=n  (n+1i)   i=1 
 Two copies, one red and the other, reversed, in green 
2 T(n)  =  i=n  (i +n+1i)   i=1 
 pair off the terms, a red with a green 
2 T(n)  =  i=n  (n+1)   i=1 
 n copies of (n+1): the i does not appear in the formula so all the terms are the same 
2 T(n)  =  n (n+1) 
T(n)  =  n (n+1) / 2 
Back to the Runsums Results page© 2003 Dr Ron Knott   12 February 2003 