NOTES

Formulas for patterns

[-Figurate Numbers-] [-Finding the formula - Finite Differences-]

Useful Formulas

We have already looked at formulas for finding the sum of arithmetic sequences (see notes on summing sequences ). Here are some more formulas which are useful in problem solving.

The sum of the first n integers 1 + 2 + 3 + ... + n = n(n + 1)/2 . ( This is also the formula for finding the n^th term ( t_n ) of the triangular numbers ).

The sum of the first n squares 1 + 2² + 3² + ... + n² = n( n + 1 )( 2n + 1 )/6

The sum of the first n cubes 1 + 2³ + 3³ + ... + n³ = n²( n + 1 )²/4

The sum of the first n powers of 2 ( starting at 2⁰ = 1 ) 1 + 2 + 2² +...+ 2^n-1 = 2ⁿ - 1

Figurate Numbers

If dots are arranged in the shape of a regular polygon (e.g. equilateral triangle, square, pentagonal etc ), their number is called a figurate number.

Triangular numbers - 1 , 3 , 6 , 10 , ...

                                       o
                      o             o  o
         o         o  o         o  o  o
o     o  o     o  o  o     o  o  o  o
1       3          6               10

Square numbers - 1 , 4 , 9 , 16 , 25 , ...

                                 o  o  o  o
                  o  o  o     o  o  o  o
       o  o     o  o  o     o  o  o  o
o     o  o     o  o  o     o  o  o  o
1       4          9               16

Finding the formulas- Finite Differences

It is trivial to find the n^th term of the square numbers ( the 10^th term is 10² = 100 !) but it is not so clear how to get the formula for the n^th term of the triangular numbers.
There is a useful method called "finite differences" which we can use and I will try to explain how it works below for you. It is somewhat complex so bear with me as it is also pretty powerful (and nifty!).

We start by placing the data in a table. Lets use the data for the triangular numbers. Notice that the two columns contain the term number (n) and the corresponding value for the n^th term ( t_n ).

term (n)	n^th term (t_n)
1	1

2	3

3	6

4	10

We now add a third column called the "first differences". In this column we place the difference between each consecutive number in the sequence. For example the first two terms are 1 and 3 so the difference is 2 ; the next pair of terms are 3 and 6 with a difference of 3.

term (n)	n^th term (t_n)	1^st diff.
1	1
		2
2	3
		3
3	6
		4
4	10

You should stop at this stage and check the values in the "first differences" column. If they happen to be a constant ( i.e. all of them the same - 1 or 23 or something ) then the relationship between the values is linear. That is the formula is of the form y = mx+c. The numbers in our pattern aren't constant so we go to the next step.
Next we add a fourth column called the "second differences". In this column we place the difference between each consecutive number in the "first differences". For example the first two values are 2 and 3 so the difference is 1 ; the next pair of values are 3 and 4 with a difference of 1 also.

term (n)	n^th term (t_n)	1^st diff.	2^nd diff.
1	1
		2
2	3		1
		3
3	6		1
		4
4	10

By now you have noticed that all the values in the "second differences" column are a constant ( 1 ). Whenever this happens we know that the relationship is quadratic . That is the formula is of the form
y = ax²+bx+c .
It only remains for us to find the quadratic equation. We do this by extending the table in the following way:

1. First of all we add another column which contains some multiple of t_n. Which multiple of t_n we put in this column depends on the constant in the "second differences" column. If the constant is 2 we simply repeat each value in the t_n column. If the constant isn't 2, we work out what we have to multiply the constant by to get the value of 2. In our example, the constant in the "second differences" column is 1 so we have to multiply by 2. So in our new column we put 2 times each of the values in the t_n column. This is shown below:

term (n)	n^th term (t_n)	1^st diff.	2^nd diff.	2t_n
1	1			2
		2
2	3		1	6
		3
3	6		1	12
		4
4	10			20

2. Next we add another column which is the result of the difference between the 2t_n column and n²:

term (n)	n^th term (t_n)	1^st diff.	2^nd diff.	2t_n	2t_n - n²
1	1			2	1
		2
2	3		1	6	2
		3
3	6		1	12	3
		4
4	10			20	4

3. At this stage we would do our differences again to find the linear equation equal to 2t_n - n². But with this example we can see that the values in the 2t_n - n² column are identical to the values in the (n) column. This means that:
2t_n - n²= n
<=> 2t_n = n² + n
<=> t_n = ( n² + n )/2
And this is the formula we wanted to find!

For your interest, if it takes three differences to get a constant than the relationship is a cubic of the form:
y = ax³ + bx² + cx + d. If the constant is equal to 6 the value of t_n is unchanged. You continue to use the process above to find the equation.

BACK