Complex Sequences

Currently in Mr.A's class you have studied the behavior and formulas of geometric and arithmetic sequences. Many sequences however don't follow either of those two patterns, in this post I will be trying to help you understand how to come up with the formula of some sequences that don't follow those patterns.

Example 1
1, 3, 6, 10, 15, 21, 28, 36, 45, 55....

As you may see, this sequence doesnt have a common number being added to or multiplied to get the next number in the sequence.

Step1
The first step to finding an explicit formula for the sequence is to find the difference between each number in the sequence, andthen finding the difference of the differences, until the number you come up with is the same.
In our example it would work like this:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55
2 3 4 5 6 7 8 9 10
1 1 1 1 1 1 1 1

Step 2
The second step is to identify what the number of rows mean. The numbers on the second row weren't the same, so we have to repeat the process. On the third row the numbers are all the same (1), meaning that we have achieved our goal. Now, what does this all mean? Well, the number of rows of differences is the degree of the equation. In this case we have two rows, meaning its a second degree equation (ax^2 + bx + c, were a,b,c are constants).

Step3
Believe it or not we have all the information we need, now we just need to make some equations and solve for the variables a, b and c. On the formulas were about to create, X will represent term in the sequence, so when we plug 1 into x the answer we need to obtain is 1, when we plug 2 into X the answer we need to obtain is 3, etc.

Therefore,

ax^2 + bx + c = 1 when x=1, so a + b + c = 1 (1)
ax^2 + bx + c = 3 when x= 2, so 4a + 2b + c = 3 (2)
ax^2 + bx + c = 6 when x = 3 so 9a + 3b + c = 6 (3)

Step 4
Now we have three equations with threevraiables, we only need to solve for the variables. The method you use is completely optional, I prefer to use elimination.

a + b + c = 1 (1)
4a + 2b + c = 3 (2) Multiplying equation (1) by (-1) and adding we obtain

3a + b = 2 (1-2) We will use this equation later.

a + b + c = 1 (1)
9a + 3b + c = 6 (3) Multiplyin equation (1) by (-1) and adding we obtain

8a + 2b = 5 (1-3) Using this equation and equation (1-2) we can solve for one variable

3a + b = 2 (1-2)
8a + 2b = 5 (1-3) Multiplying equation (1-2) by (-2) and adding we obtain

2a = 1
a= 1/2

Plugging a into (1-3)
8(1/2) + 2b = 5
2b = 1
b = 1/2

Plugging a and b into (1)
1/2 + 1/2 + c = 1
c= 0

Replacing all the values for the constats our final equation for our sequence is:

(x^2)/2 + x/2 = nth term of the sequence

Simplifying:
(x^2 + x)/2 = nth term

Factoring an X we obtain the final equation:
(x)(x+1)/2 = nth term

This is the equation to find the sum of the first x integers. Which, if you refer to the original sequence, you will notice that that was the pattern the sequence was following.We can now test this equation to find the sum of the first 8 integers.
(8)(8+1)/2) = 36
1+2+3+4+5+6+7+8 = 36 so the equation works.

Example Problem:
Now that you know the steps to find an explicit equation, you should try to find the equation for the sum of the first n integers squared. The first terms of the sequence are:

1, 5, 14, 30, 55, 91...