Proving that 1+2+3+...+n is n(n+1)/2
We give three proofs here that the n-th 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 a
rectangle. This is followed by two proofs using algebra. The first uses "..." notation and the second introduces you to the
Sigma 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 |
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... | |
|---|
| 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 n
we 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 upside-down 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 + | ... + | (n-1) + | n |
| + | n + | (n-1) + | (n-2) + | ... + | 2 + | 1 |
Two copies, one red and the other, reversed, in green |
| = | (1 + n) + | (2 + n-1) + | (3 + n-2) + | ... + | (n-1 + 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 pair-sums 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 letter
and 
(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 the
mathematical symbol for integration - another kind of sum).
Mathematicians use the capital sigma for the sum of a series as follows:
- a formula describes the ith 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 ith term) |
|
| i=starting value |
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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 which
is 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 102+112+122, where the numbers added are the square numbers i2: |
| i=12 | i2 |
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| 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 |
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| 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:
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| n | i = T(n) |
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| i=1 |
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and this time, we have T(n) but written backwards:
n+ (n-1) + ... 3 + 2 + 1
where the ith term is now n+1–i for i from 1 to n:
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| i=n | (n+1–i) = T(n) |
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| i=1 |
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Finally, note that if all the terms are independent of the variable, for instance
if 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 be
given 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 |
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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+1–i) |
|
| i=1 |
|
Two copies, one red and the other, reversed, in green |
| 2 T(n) |
= |
| i=n | (i +n+1–i) |
|
| 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 |
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