Bam! Mr. Tarrou. In this video we are going to continue our discussion of sequences in
PreCalculus with an introduction of Arithmetic Sequences. It is a special type and we are
going to learn about 2 types in the next couple of videos. An arithmetic sequence is a sequence
in which each term after the first term differs from the preceding by a constant amount or
difference. They are very very easy to spot this pattern. They show up so often that we
are going to end this initial board by discussing a formula to help you set up the general term
or the general form of an Arithmetic Sequence. But first lets just take a look at this example
of A sub 1, A2, A3, and A sub 4. We have the values of 2, 11, 20, and 29. Let's just look
at what we have got there. A sub 1, the first term in our sequence is equal to 2. And d,
our common difference, what are we constant adding or subtracting by.... 11-2 is 9, 20-11
is 9, and 29-20 is 9. So we have a list of numbers, a sequence of numbers and this is
written as finite because I don't have ... at the end indicating that it is going continuously...
where you recognize that as you subtract each pair of numbers like A4-A3, A3-A2, and so
on that you continually get this common difference of 9. So when you find a sequence of values,
you are constantly adding or subtracting by the same number, you are dealing with an Arithmetic
Sequence. We are going to come up with two formulas now, two versions of... not versions...
but two formulas or expressions that will create the numbers in this sequence for us.
The first type is the Recursion formula. We have talked about this in the previous video.
It is a pattern, it is a combination of two formulas. Recursion formulas are very easy
to come up with. Their only problem is, if you want the tenth term you have to start
at the first one and work through all of the first nine before you get to that tenth term
and you figure it out. Then we are going to talk about the General term which will allow
us to go right to that tenth term or any term in the sequence we like skipping all of the
preceding values or terms. A recursion formula, well recursion formulas have two parts. There
is an A sub 1 formula and then there is an An formula. So a recursion formula... Just
look at what we have there. The first term in this sequence is equal to 2, so A1 is equal
to 2. The recursion formula allows you to define every term in the sequence by doing
something with the preceding term. Like how do I find 11? I add 9 to 2. How do I find
20? I add 9 to 11. How do I find 29? So I am simply just adding by 9. So my second formula
in the Recursion formula is going to be An equals A sub n-1... notation to say you want
the third term, go the second term... If you want the fifth term, go to the fourth one.
Just go the preceding term in the sequence and then add 9 to it and you are done. I started
with 2 and I added by 9, not 11. Ok. Done! But again, if I want the tenth term with this
set up I have to go through all of the preceding 9 terms. The General term. Now this is a big
deal. This... You will get a similar formula when we get into the Geometric Sequences.
This is a very important formula for Arithmetic Sequences. It is what allows you to quickly
set up an expression to get to any term in the sequence that you like. Notice that this
equation has no place in that asks for A sub n-1. There is n-1, but it is not a subscript.
It is not telling you where you are in a list. This General term of an Arithmetic Sequence
will allow you to go to any term in that sequence in one single step. The general term is An,
we are going to describe this arithmetic sequence with A sub n equals A1, so we are going to
start with that first term again, plus n-1 times d. We are adding by the common difference,
which could be negative so you are adding by a negative, and we are going to add by
that common difference n-1 times. Let's just look at this for a second before we move on
and write the answer. We are at 2. How many times do we have to add to get to 11, the
second term. We only add by 9 once. To get to the third term, we only had to add twice.
To get the fourth term we only have to add three times. I teach on the third floor and
students are constantly saying, "Oh those three flights of stairs wore me out." You
did not take three flights of stairs, you took 2 flights. You were already on the first
floor. We are looking at A1, so when you add by 9 or whatever your common difference is,
you are not getting to your first term you are getting to your second term. Thus to get
to your fourth term, you only have to add three times. Just like to get the third floor
of my building you only have to walk up 2 flights of stairs because you are already
on the first floor. So that is where this n-1 comes from and our general term is going
to be An equals A1 which is 2 plus n-1... again in a sequence as well n is a positive
integer as the domain. It is just telling you where you are in a sequence... times the
difference of 9. And it would be nice, and your teacher may want like I do, will want
you to simplify this expression. We have 2+9n-9. And 2-9 is negative seven. We have An=-7+9n.
Alright. Let's go look at a couple more examples. Write the general formula for an arithmetic
sequence with a first term of 14 and a common difference of -3. Very similar to the previous
example except this is written out in sentences without giving you the actual numbers. So
we are told that it is arithmetic. So as soon as we are told that a sequence is arithmetic
we can go ahead and write that general form out An=A1+(n-1)d. And let's identify the important
information and just simply write our answer. The first term of the sequence is 14. So A1=14.
And the common difference is -3, so d=-3. We are actually ready to write this answer.
A1 is again equal to 14... plus n-1... and our common difference is -3. We will put the
negative three in here. I get students that write their answers like this very commonly,
where it is A1... n-1... and there is the d which is -3. But look at what I wrote. That
does not say n-1 times d, that says n-1 minus 3. Let's make sure that we wrap that common
difference in a set of parenthesis. Because my book, all of the books that I have used
for Arithmetic and Geometric sequences... we are doing arithmetic right now... n-1 is
always written as the coefficient of d. It might make more sense actually if you were
to flip this around so students would not make this sign error or notation error. We
are going to then take this -3 and distribute it through the parenthesis just to clean this
up a little bit. We have 14... -3 times n is -3n... -3 times -1 is positive 3. Now you
might noticed that I dropped the parenthesis there. There is nothing in front of these
parenthesis here except a plus sign. I could take that extra step of writing plus and then
the negative 3n and the positive 3. But distributing positive one through parenthesis is not going
to change any values. So it is really just there to highlight the factor of n-1 and that
positive one in front of the parenthesis is not going to change anything as it goes through
there. So really these... Once you take the -3 and distribute it through the parenthesis,
they don't need to be there anymore. And we have like terms of 14 and 3. We have -3n+17.
That is our An notation or our general term of this arithmetic sequence. Let's take a
look at one more example and we will be done. Last example. Express the sum using summation
notation. So I hit these a lot when we go over our notes because they are the most thought
provoking problems in these sections I am covering right now in PreCalculus. You cannot
just completely rely on memorization of formulas that help you out with some of these questions
because you are trying to come up with formulas based on patterns. That can be a little bit
difficult. So we are going to write a summation notation for this. And we are going to have
a lower index... We need a lower index and an upper index. Of course ultimately we need
some kind of expression. That clearly is going to be a fraction that will create the numbers
that our in our sequence of addition that we are going to make this summation notation
with. Ok. Let's draw our attention to whatever pattern that we see first. The denominator
is increasing by values of 3. Now you might save that easier pattern for the last step.
You might want to work on the numerator first. But, I am going to focus on the first pattern
that I see which is in the denominator. I am seeing -5, -2, 1, 4, and 7 all the way
up to 19. I am seeing an Arithmetic sequence in the denominator of my values here. So I
am going to use this formula, this general term ,of an arithmetic sequence to come up
with an expression that will describe the denominator of these fractions. An is equal
to A1 which is -5, and my value of d... I am constantly adding by the value of 3 so
my constant difference is 3. We will go ahead and distribute this to clean it up a bit.
We have -5+3x-3. Combine like terms, we have a negative five and a negative three. That
is -8+3n. Sorry about the squeaky chalk. So the denominator of my fraction is going to
be -8+3n. And when you set up your summation notation you are allowed to have your index,
which is going to be described with k here, you are allowed to start your index with any
integer you like. You can start this with -2, 0, 1, 5, whatever you like. But since
I am using my An notation or my general term formula for Arithmetic Sequences your domain
here for n values are all positive integers. The lowest, first value of n that I am going
to use if I am going to use this formula to help me figure out this denominator is 1.
So I am going to start k with a value of 1 and let it go up by 1. 1, 2, 3, 4, 5, all
the way up to... I don't know what that last n value is that I am going to use for this
sequence of addition. So I am going to start my lower index. But now I have to figure out
what is my upper limit. So to do that I am going to erase this and write my An formula
out. So A sub n is equal to -8+3n. My last An value... This just tells you where you
are in a sequence, but An is the term. My last term is 19. So I am going to take out
An and plug in nananana...19 and break my chalk. Ok. What do we have here. We are going
to take the -8 and add 8 to both sides. 19 and 8 make 27. Divide both sides by 3. I don't
know why I changed colors there. 9 is equal to n. So my upper limit that I am going to
use in my summation notation is 9. I am really getting close to being done with this expression.
What is going on with the numerator? In the numerator I am multiplying by 1 to get to
1. Then I am multiplying by 2 to get from 1 to 2. Then I am multiplying by 3, then I
am multiplying by 4, and so on. So I have a pattern in the numerator that is multiplication.
My multiplication values are increasing by 1. That is just the operation of the factorial
symbol. 5! or 5 factorial is 5 5 times 4 times 3 times 2 times 1. So when you see that multiplication
pattern changing by values of 1, or decrease by one as we go to the left, that is your
clue that this is a factorial pattern going on in the numerator. So let's see if we can
just put k factor. I don't think so. 1 factorial is 1. Ok, that checked out. 2! though is not
equal to 1. 2!, because we are going to put k! or some form of k! in the numerator, 2
factorial is equal to 2. 3! is equal to 6. 4! is equal to 24. So this is 1 factorial,
this is 2 factorial, so when k is 2 we are doing 1!. When k is 3 we are doing 2!. When
k is 4 we are doing 3!. And when k is 1 we are doing 0! which by definition is equal
to 1. So this pattern is not k!, but it is (k-1)! I am Mr. Tarrou. BAM!!! Go Do Your
Homework:) NOOOO!!!!! I need to get my red chalk out but I know the camera would not
pick up very well. Did anybody noticed that I accidentally went into the variable n and
not the variable k. So everything else I said was right. But I checked my work after turning
the camera off and I copied the wrong letter. Now as you do your homework, check your work
too. BAM!