➕➖✖️➗ Sequences and Series

Welcome to lesson 1 of this new unit, Sequences and Series. In this new unit, we will go over explicit formulas, understand how to solve them, and understand what they mean.

Let’s start with what sequences and explicit formulas are.

Sequences
Definition: A series of numbers in a list separated by a to-be-determined repeating amount
Example: 1,2,3,4,5,6,7,8,9 Here, the list is separated by +1


Explicit Formulas
Definition: An explicit formula calculates the unknown term in a sequence



Term
Definition: an individual number in the sequence
Example: 1,2,3,4,5,6,7,8,9 Here, a possible term is 5

Variables used
n: A lowercase n variable is used to label the number in a sequence

t: A lowercase t variable is used to label the term

Welcome to lesson two of our series and sequences unit. In this lesson, we will be going over the recursion formula and some examples of the recursion formula.

Let's start with understanding the recursion formula and the variables we will use for this unit.

Recursion Formula:
A recursive formula uses the last or set of previous terms to define each term in a sequence.

Variables:
Yes, any variables that you find easy to use or comfortable with you are free to use but for this lesson, I will be using the variables t and n

Just like the last lesson, these variables mean:
n: A lowercase n variable is used to label the number in a sequence

t: A lowercase t variable is used to label the term

Welcome to lesson three of the series and sequences unit. In this lesson, we will be discussing Pascal’s Triangle. Let’s start by exploring and understanding the terms and definitions we will use in today’s lesson.

Let's start with understanding what the pascal triangle is.

The Pascal Triangle:
To create a triangle consisting of a border of ones and adding the inside of each row

Example of a Pascal Triangle:


1
1 + 1
1 + 2 + 1
1 + 3 + 3 + 1
1 + 4 + 6 + 4 + 1
And so on…………

The variable we are going to be using in todats lesson are

n:
The variable n represents the row number and each triangle starts at row 0

r:
The variable r represents the column number and each triangle starts at column 0

t:
The t represents the term

Example:
t2,3 =

Welcome to lesson 4 of the sequences and series unit. In This lesson, we will be going over the Arithmetic Sequences.

Let's start with understanding what Arithmetic sequences actually are and the lesson and formulas needed in this lesson

Arithmetic Sequences:
Arithmetic sequences are a series of numbers which are increasing or decreasing by the same number each time.

Variables used:
tn:
term number of the sequence

n:
The number in a series

a:
The first term in a sequence is represented by the variable “a”
For example 1,2,3,4,5,6,7,8
In this series of numbers, the “a” would be 1 as it's the first term.

d:
The variable “d” is what the series of numbers is increasing or decreasing by.
Example 2,4,6,8,10

a = 2
d= 2 as the series is increasing by two each time so the “d” value is positive 2

Welcome to lesson 5 of the sequences and series unit. In this lesson we will be going over Geometric sequences. As always lets start this lesson with understanding what we are learning.

Geometric sequences:
A geometric sequence is a series of numbers in which each term is found by multiplying the previous term by a constant factor

Variables used

tn = is the term number and n is the certain number in a series of numbers

a:
The variable “a” is the first term in a series of numbers

r:
The variables “r” is the constant factor in which the term is bing multiplied by each time.

Welcome to lesson 6, our final lesson on the Series and Sequences unit. In our final unit we will be learning different ways to solve Arithmetic and Geometric series questions

Variables needed Arithmetic Follow The dotted lines

Variables need Geometric follow the solid lines

In geometric sequences a few formulas are needed to solve some problems.

General formula: helps find the nth term from a series of numbers

tn = a * rn-1

Recursion formula:

tn = r * tn-1

Steps

Step 1

Locate the variables given to you in the question

Step 2

Input the found variables into the correct formula

Step 3

Solve for the necessary missing variable

EXAMPLE PROBLEM

Find t40 with the sequence 7776, 1296, 216, 36, 6

Step 1

a = 7776
r = divided by 6 but since only multiplication x1/6

Step 2

We will use the general formula as we are finding the nth term

tn = a * rn-1

t(40) = 7776 * 1/640-1

Step 3
Solve

t(40) = 1/1296

All terms can be represented by tn,r and this coils be represented by the formula

tn,r = tn-1,r-1 + tn-1,r

Steps

Step 1

Start off by drawing the pascal triangle. You can go up to the row needed or add some extra just in case.

Step 2

Based on what the question is asking locate the rows and column you will be needing in this problem

Step 3

Solve the question

EXAMPLE PROBLEM 1 Find the value of these coordinates using the Pascal Triangle.

a) t3,3

Step 1
The question is asking for the 3rd row
And the 3rd column so start with making a
Pascal triangle with these requirements

1
1 + 1
1 + 2 + 1
1 + 3 + 3 + 1


Step 2
using the created Triangle we can locate
Row 3 and column 3 in that row

Step 3
Now we can solve
t3,4 = 1

b) t6,2

Step 1


The question is asking for the 6th row
And the 2nd column so start with making a Pascal triangle with these requirements
Step 2
using the created Triangle we can locate using the now made triangle, we can
Row 6 and column 2 and locate the row and column in the question.


1
1 + 1
1 + 2 + 1
1 + 3 + 3 + 1
1 + 4 + 6 + 4 + 1
1 + 5 + 10 + 10 + 5 + 1
1 + 6 + 15 + 20 + 15 + 6 + 1

Step 3

Now we can solve t6,2 = 15

EXAMPLE PROBLEM 2 Expand and simplify

(x + y)6

Step 1

1
1 + 1
1 + 2 + 1
1 + 3 + 3 + 1
1 + 4 + 6 + 4 + 1
1 + 5 + 10 + 10 + 5 + 1
1 + 6 + 15 + 20 + 15 + 6 + 1

Step 2

Now we can see that to expand we will be needing row 6’s coefficients

Step 3

(x + y) 6
= 1 x6y0 + 6 x5y1 + 15 x4y2 + 20 x3y3 + 15 x2y4 + 6 x1y5 + 1 x0y6

Steps:

Step 1

Look for the t value or any other values given to you in the problem. Some might be obvious such as t1 = 2, while others might be harder to find.

Step 2

Once the given value is found, there will also usually be either a recursion formula or a constant the table changes which is what you need to find next.

Step 3

Once you have the formula, either by making it or giving it to you, you need to answer the question by implementing the values into the formula.

EXAMPLE PROBLEM ONE: Solve for the first three terms
t1 = 6

Formula:
tn = tn-1 -12



Step 1
In these questions, we are given the first t value of t1 = 6

Step 2
It is also known that we are given the formula for the question, which is seen to be tn = tn-1 -12

Step 3

Solve the necessary question.

t(2) = t2-1-12
t(2) = t1 -12
*here, we can see that t1 reapers into the equation, and knowing what t1 is since it was given ot us, we can input that into the equation.*

t(2) = 6-12
t(2) = -6

*Term 2 is -6*

Repeat step 3 for the rest of the terms

t(3) = t3-1-12
t(3) = t2 -12
t(3) = -6-12
t(3) = -18

t(4) = t4-1-12
t(4) = t3-12
t(4) = -18-12
t(4) = -30

EXAMPLE PROBLEM TWO
There are 15 rows in a movie theatre. The first row has twelve, the second row has 24, and the third has 36. How many seats does the third last row have?




Step1

Here seeing that the first three rows are given we are able to gather all of those t values

t1=12
t2=24
t3=36

Step2

Finding the formula
t(n) =(nx12)+12
We known from the first three terms that the sequence is going up by 12 and that the adding 12 to the previous value will give us the next hence adding 12 to 12 time to the number

Step 3
Solve

Now that we have the t value and the formula we can solve for what the question is asking for

t(12) = (12x12)+12
t(12) = 144 + 12
t(12) = 156


The third last row has 156 seats

EXAMPLE PROBLEM THREE solve for the first 2 terms
t1 = 56
Formula:
tn = 82(tn-1)

t(2)= 82(t2-1)
t(2) = 82(56)
t(2) = 4,592

t(3) = 82( t3-1)
t(3) = 82(t2)
t(3) = 82(4,592)
t(3) = 376,544

Formulas:

sn = n[2a + (n-1)d] / 2

And

sn = a * (rn-1)/r-1

*Also look at the yellow and dotted green lines *

Steps

Step 1

Locate the variables given to you in the question

Step 2

Input the found variables into the correct formula

Step 3

Solve for the necessary missing variable

EXAMPLE PROBLEM


Find the sum of this geometric series using this sequence 1/49 + 1/7 + 1 + 7+..............+40353607.

Step 1

Sn =?
n = ?
a = 1/49
r = 7

Step 2

Since we have two variables missing we need to use the general formula to solve for the missing n variable

Tn = a * rn-1

40353607 = (1/49)7N-1

Step 3

40353607 / 1/49 = 7n-1

1977326743 = 7n-1

711 = 7 n-1

11 = n-1
10 = n

Now that you only have a single missing variable you can solve for it.

Sn = 1/49 (710-1)/ 7-1

Sn = 960800.1633

General formula which helps find the nth term:


tn = a + (n-1)d

The recursion formula is a formula which defines any term in a sequence using previous terms

tn = tn-1 + d

General formula which helps find the nth term:


tn = a + (n-1)d

The recursion formula is a formula which defines any term in a sequence using previous terms

tn = tn-1 + d

Steps

Step 1

Locate the variables given to you in the question

Step 2

Input the found variables into the correct formula

Step 3

Solve for the necessary missing variable

EXAMPLE PROBLEM 1
Find t20 using the sequence 90,80,70,60,50,40

Step 1

Find the given variables

a = 90
d = -10
n = 20
t(20) = ?

Step 2
Since we are solving for an unknown term we can use the general term

tn = a + (n-1)d

t(20) = 90 + (20-1)(-10)

Step 3
Solve for the missing variable

t(20) = 90 + (19)(-10)
t(20) = 90 + (-190)
t(20) = -100

Steps to solve

Start with drawing a table or any other recognition technique to help you understand what the sequence is adding or subtracting by
Understand and uncover what the sequence is changing and check to see if it matches the other numbers.
Next, create a fundamental equation for the sequence using tn = n and adding the variable of which the code is increasing; by for example, if the sequence was 2,4,6,8 since the series is going up by two each time, a possible equation could be t(n)=2n

Finally, a check could be helpful to see if the equation you created works and is correct. The check code is done by taking your equation, adding one of the starting terms from the given sequences into your code, and solving it to see if the answer is correct.

EXAMPLE PROBLEM ONE: Find the formula for the nth term

8,9,10,12,14

Step 1:
n t(n)
1 8
2 10
3 12
4 14



Step 2(could be mentally but should be thought of):

The sequence goes up by two each time.

Step 3:
Base Formula
t(n) = n

Changed for function:

t(n) = 2n+6

Step 4:


Check:
We can check our formula using one of the already given values from the table to see if we get the correct one. t(n)

Start with writing the formula you created and implementing a value from the table

t(2) = 2(2)+6
t(2) = 4+6
t(2) = 10
= The formula checks out

EXAMPLE PROBLEM TWO: Given the sequence, find the 1st, 3rd and 7th term using the formula tn = 5n + 7

Here we are given a formula and are told to find the first, third and seven.

Start with replacing the n value with the # and solve the rest.



a) tn = 5(1) + 7 b) tn = 5(3)+7 c) tn = 5(7)+7
tn = 5+7 tn = 15+7 tn = 35+7
tn = 12 tn = 22 tn = 42

^

EXAMPLE PROBLEM THREE

To help us understand the actual usefulness of explicit formulas, we can do some higher calculations with the formulas. Find t55 and t89 using the following sequence of 8,16,24,32,40

n tn
1 8
2 16
3 24
4 32
5 40


Formula is:
tn = 8n

Check:
tn = 8(4)
tn = 32


a)
t(55):
t(55) = 8(55)
t(55) = 440

b)
t(89) = 8(89)
t(89) = 712