Main Series Page

- Sum to n of the Squares of Natural Numbers Using Differences
- Sum of the squares of the first n natural numbers using summation
- Sum of Natural Numbers Squared Using Errors
- Using Infinite Calculus to find the Sum of the Squares of the First n Natural Numbers
- Summing the first n natural numbers by finding a general term

n | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|---|

n^{2} |
0 | 1 | 4 | 9 | 16 | 25 | 36 |

S_{n} |
0 | 1 | 5 | 14 | 30 | 55 | 91 |

Δ_{1} |
1 | 4 | 9 | 16 | 25 | 36 | |

Δ_{2} |
3 | 5 | 7 | 9 | 11 | ||

Δ_{3} |
2 | 2 | 2 | 2 |

As usual, the first n in the table is zero, which isn't a natural number.

Because Δ

an

and we can find the coefficients using simultaneous equations, which we can make as we wish, as we know how to add squares to the table and to sum them, even if we don't know the formula.

In the table below, we create three equations, noting that d=0 from the first one (revealing the reason for the non-natural number zero)

n | 0 | 1 | 2 | 3 |
---|---|---|---|---|

S_{n} |
0 | 1 | 5 | 14 |

d=0 | a+b+c=1 | 8a+4b+2c=5 | 27a+9b+3c=14 |

Rewriting our equations:

a+b+c=1 [1.1]

8a+4b+2c=5 [1.2]

27a+9b+3c=14 [1.3]

Using 1.1 with 1.2 and 1.3 we can make two new equations:

6a+2b=3 [1.4]

24a+6b=11 [1.5]

By subtracting 3x Equation 1.4 from 1.5, we get:

6a=2

So a=1/3 [Also noting, by the way, that Δ

Substituting a=1/3 in 1.4 gives

2b=1

So b=1/2

Finally, substituting these values into 1.1 we get

1/3+1/2+c=1

So c=1/6

[Actually, with the sum of the powers, the sum of the coefficients in the formula is always 1]

So, we can substitute our values into 1.0, to get the sum of the squares of the first n natural numbers (or first n positive integers):

n

Or, in various forms:

As before, we set up as follows:

Saying the sum to n is one term less than the sum to (n+1)

We expand the sums:

As expected, the cubic terms cancel, and we rearrange the formula to have the sum of the squares on the left:

Expanding the cube and summing the sums:

Adding like terms:

Dividing throughout by 3 gives us the formula for the sum of the squares:

Or:

The graph below of y=x

By integrating the x

[3.1]

Where is the sum of the squares and E

The sum of the n-1 squares and the error E

[3.2]

We note that the difference between 3.1 and 3.2 is;

[3.3]

being the nth square

The difference between 3.1 and 3.2 is also:

[3.4]

Expanding (n-1)

[3.5]

Rounding up similar terms:

[3.6]

From 3.3, we get:

[3.7]

Expressing the errors in terms of n:

[3.8]

This is a recursion formula. If we know one of the errors we can find the other. We do know what E

E

Letting n=2, and changing the relationship a bit:

E

E

Similarly,

E

Or:

E

So we can generalise, the k-th term is:

[3.9]

And the nth term is:

[3.10]

We know the sum of the first n natural numbers,

()

so we can calculate (write down) the summation in 3.10 and put the result in

[3.11]

Simplifying:

[3.12]

Remembering:

[3.13]

Substituting 3.12 in 3.13, we get the formula for the sum of the squares of the first n natural numbers:

[3.14]

The following graph is of y=x

y=x2 represents part of the sum of the squares, and the rest is the area between each rectangle and the function.

Looking at the graph, it seems the difference, or error E

The sum of the squares, where E

[4.1]

The error is the area of each square less the area under the graph:

[4.2]

Integrating:

[4.3]

Expanding terms:

Collecting like terms:

[4.4]

And summing:

[4.5]

The error is:

[4.6]

Substituting 4.6 for E

[4.7]

Set up for generalising:

[5.1]

Beginning to unwind:

[5.2]

Continuing to unwind:

[5.3]

Writing the nth term:

[5.4]

(nn

It is now evident that the sum of the squares on both sides will cancel out. We could get a sum for the natural numbers out of this, but we can do that directly and much more easily.

After some thought, we can conclude that the last term must be negative for us to find an expression for the sum of the squares. It was negative for the sum of the natural numbers, and it will be negative again for the sum of the cubes, but will not work for the sum of the powers of four. That is, this method works for the squares of the odd numbers.

The expression corresponding to 5.4 for the sum of the cubes is:

[5.5]

Here the sum of the cubes on the right-hand-side is negative. I also note that in 5.4 and 5.5 the coefficients correspond to the binomial coefficients of the square and the cubic

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