We have already used Gauss's trick to sum the natural numbers.
Sum to n of the Natural Numbers Using Differences
We can find the formula for a set of numbers using differences. For instance, the following table shows the sum of some natural numbers, but we have also used zero, for convenience:
n
0
1
2
3
4
5
S_{n}
0
1
3
6
10
15
Δ_{1}
1
2
3
4
5
Δ_{2}
1
1
1
1
S_{n} is the sum of the numbers to n.
Because we find that Δ_{2} produces constant values, we assume the formula for the sum of the natural numbers is a quadratic, of the form an^{2}+bn+c.
Using our values, we substitute 0, 1, and 3 in the Equation:
n
Equation
Equation Number
0
c=0
1
1
a+b=1
2
2
4a+2b=3
3
In Equations 2 and 3, we have noted that c=0. By subtracting twice Equation 2 from Equation 3, we get: 2a=1, So a=1/2 Substituting the value for a in Equation 2, we find that b is also 1/2, So the sum of the first n natural numbers, S_{n}, [As
a word to the wise, the constant value in the table above is always
(n!)a, so in the example, a=1/2!, or 1/2. The sum of the coefficients
of the sum of the powers of the natural numbers is always 1.]
Sum of the natural numbers using summation
In this scheme, we first fall flat on our faces. We note the relationship:
That is, the sum of n natural numbers is the sum of n+1 natural numbers less (n+1). Expanding the term for (k+1), we get:
Which simplifies to:
We
can try another approach, and look for the sum of the squares of the
first n natural numbers, hoping that this sum will vanish.
Second Try With Summation
Starting again, we note that the sum of the squares of the first n natural numbers is the sum of the first (n+1), less (n+1)^{2}.
Expanding the (k+1)th term:
Expanding (n+1)^{2}, and rounding up similar terms:
Which gives us the sum of the first n natural numbers:
The following graph is of y=x, and the rectangles represent the natural numbers 1, 2, 3, 4:
The area of the triangular graph, if we take it as representing n numbers is: n^{2}/2 So the sum to n terms is, approximately:
If we take E_{n} to be the error on the approximation to n terms:
Looking at the graph, the error on each number is 1/2, so the error on the sum of the first n numbers, E_{n}, is n/2. So:
Second Approach With Errors
As we cannot figure out the error this easily every time, we try another approach: [3.1] And the error on S_{n1}: [3.2] Noting the difference between the two sum, in preparation for subtracting 3.2 from 3.1:
[3.3] Subtracting 3.2 from 3.1, taking into account 3.3:
[3.4] Simplifying and rearranging: [3.5] Normally,
this would be the start of things, but in this case, the error is
simply a number, independent of the actual nvalue, so it is the same
for all n, and the error is n/2, as we found above.
Using Infinite Calculus to find the Sum of the first n Natural Numbers
This
approach is similar to the previous one, but introduces a different
approach that can be used with other natural number sums.
In the graph below, we have the first few natural numbers shown as rectangles on the graph y=x.
The area under the graph approximates the sum of the natural numbers, so: [3.21] With E_{n} as the error on S_{n}: [3.22] Each of our rectangles in the graph have an area k∙1, so our error is k minus the area under the graph between (x=k1 and x=k): [3.23] Working out the integral, we get: [3.24] Replacing the integral in 3.23: [3.25] Simplifying the sum, adding it up:
So substituting our value for E_{n} in 3.22 we get:
Which comes as no surprise!
Summing the first n natural numbers by finding a general term
Here we unravel the righthand side and hope to find a formula. We note the difference between the sum of the first n natural numbers, and the sum to (n1) is n
The idea is that we look at the terms S_{n}S_{n2}, et c, and write down the know differences, hoping that a pattern appears and we can write S_{nk}, and then to write down the nth term, from which we can extract a formula.
And for the sum to (n2):
Continuing to unravel:
And one more:
A pattern becomes clear. So we can write a general term, the kth term:
[3.3.1] We have noted that the "n" term is always kn, and the other term is a sum of the natural numbers.
Now we can make k=n, and so get our nth term, noting S_{0}=0 [3.3.2] The sum above is one n short of S_{n}: [3.3.3] Substituting 3.3.3 in 3.3.2:
Rounding up stragglers:
And finally,
Summing the First n Natural Numbers Using Number Theory
This method actually leads to a new formula for the sum of the first n natural numbers.
According to number theory, we can represent numbers in one of these four ways: [4.1] Where 4m±1 clearly represents the odd numbers.
We try to form an expression for the squares of the odd numbers by squaring 4m±1, [4.2]
Factorising part of 4.1: [4.3]
If we form this expression [4.4]
We can write the square of any odd number as: [4.5] We can write an odd number as (2n1), where n is a natural number.
n
2n1
(2n1)^{2}
8x
q
+1
8q+1
1
1
1
8x
0
+1
1
2
3
9
8x
1
+1
9
3
5
25
8x
3
+1
25
4
7
49
8x
6
+1
49
5
9
81
8x
10
+1
81
6
11
121
8x
15
+1
121
7
13
169
8x
21
+1
169
8
15
225
8x
28
+1
225
9
17
289
8x
36
+1
289
10
19
361
8x
45
+1
361
We note that we can find a q value for each of the numbers, so that 8q+1 is equal to the square of the odd number.
It
is possible that these q's look familiar in some way. After a short or
long time, of thinking about this, we might realise that the q's are
the sum of the natural numbers.
[4.6]
Substituting (2n1)^{2} for the odd number squared and 4.6 in 4.5 we get: [4.7]
Rearranging and expanding the square: [4.8]
Adding similar terms and dividing by 8, we get
[4.9]
And adding n to both sides: [4.10]
There is one issue, which we deal with in the next section: How can the following: [4.4, repeated]
be the same, in some way, as the sum of the first natural numbers?
New Formula for the Sum of the Natural Numbers
We have observed that in some way (2m^{2}±m) and (n^{2}/2+n/2) give similar results under some circumstances. As I can't see any clear relationship, I make a table:
n
(n^{2}+n)/2
m
+/
2m^{2}+/m
0
0
0
0
1
1
1

1
2
3
1
+
3
3
6
2

6
4
10
2
+
10
5
15
3

15
Naturally, the minus signs alternate and the m values occur
twice for each value. Apart from this, I see nothing except... it seems
I need to relate the n values to the m values.
Of course, I do another table:
n
0
1
2
3
4
5
m
0
1
1
2
2
3
mn
0
0
1
1
2
2
n
6
7
8
9
10
11
m
3
4
4
5
5
6
mn
3
3
4
4
5
5
(The length of the table indicates that I did not see anything quickly).
However, the following relationship comes to mind: m=ceil(n/2) [5.1]
Substituting this into 2m^{2}±m, from 4.4, gives: [5.2]
If n is even, then ceil(n/2)=n/2, and the second term will be positive, giving our normal formula.
If n is odd, [5.3]
Substituting this in 5.2, we get:
which simplifies to our normal equation for the sum of the natural numbers.
Of course, I prefer our old formula!
Application  Sum of Odd Numbers
The formula for the sum of the natural numbers can be used to solve other problems.
The sum of the first n odd natural numbers is (2k1 represents any odd number): [6.1]
Rounding up like terms, the sum of the first n odd natural numbers is: [6.4]
Application  Sum of Even Numbers
The first even numbers are (2k is always even, and represents any even number): [7.3]
Expanding the lefthand side: [7.4]
Applying our formula for the sum of the first n natural numbers: [7.5]
The
sum of the first n even numbers is bigger than the sum of the first n
odd numbers, because the first even number (2) is bigger than the first
odd number (1) and this pattern continues (4 is bigger than 3). Clearly
it is always bigger by n.
The sum of the odd number is bigger
than the sum of the natural numbers, because, after 1, the odd numbers
are bigger than the corresponding natural number (3 is bigger than 2,
and 5 is bigger than 3).
Application  Sum of Part of the Series of Natural Numbers
The sum of part of a series from n_{1} to n_{2} is: [7.5] The sum of part of the series of natural numbers from n_{1} to n_{2} is the sum from 1 to n_{2}1 less the sum from 1 to n_{2}. [7.6]
Substituting the formula for the first n natural numbers in 7.6, we get: [7.7]
Which gives us: [7.8]
Collecting like terms: [7.9]
Factorising gives us the formula for the series of natural numbers from n_{1} to n_{2}: