BAM!
Mr. Tarrou.
In this video we are going to continue our discussion about arithmetic sequences.
We are going to do that by talking about the nth partial sum.
The sum of the first n terms of an arithmetic sequence is found with the formula S sub n
equals n divided by 2 times A sub 1 plus A sub n.
Let's not forget that n is the number that you are plugging into this formula or that
tracks where you are in an arithmetic sequence.
So when n is equal to 4, you are looking at the fourth term in the sequence.
If n is equal to 5, you are looking at the fifth.
A sub n is that value that you get, that actual term... so this is A sub 1 is 2, A sub 2 is
equal to 11, A sub 3 is equal to 20.
And that is when n equals 1, n equals 2, and n equals 3.
And again, with arithmetic sequences... you recognize that by looking at your list of
numbers and observing that there is a constant addition.
You can be adding by a negative number which will look like subtraction.
But there is a constant addition.
So we are going to from 2 to 11, we are adding by 9.
From 11 to 20, we are again adding by 9.
And adding by 9 to get to 29 and so on.
Or you know... you could say it in math terms.
A sub n minus A sub n-1 is a constant difference.
So back subtract and make sure that you have the same difference for all of your terms
in the sequence.
Ok.
At any rate, let's find these partial sums.
Here we have got 2, 11, 20, 29, 38, and 47 and... you know... for space purposed I did
not write out the question but i would like the partial sum.
I want the sum of the first six numbers.
Now it is only six numbers so we could just simply add this up.
But this formula is to save time when we have to add 50, 60, 70, 80, 100 numbers or even
more.
Where you would not want to just take 47 plus 38 plus 29 and so on and just manually add
those up.
So S sub 6, we are going to find the sum of the first six terms in the sequence... that
is what this subscript means... is equal to n, or 6, divided by 2 times A1 which is 2
plus An which is 47.
Ok.
Well, you know... follow the order of operations.
We are going to 2 plus 47 makes 49.
And...
well...
Let's clean this up as well.
Six divided by two is equal to 3.
So we have 6 divided by 2 is 3.
Cleaning up inside the parenthesis we have 49.
Then this partial sum is equal to... let's see here.
3 times 40 is 120.
And 3 times 9 is 27.
so 120 plus 27 is 147.
Let's just make sure that is right.
Yes it is.
Ok.
Moving on to our second example.
Find the sum of the first 30 terms.
Now here I am going to extend this previous example where you don't actually know what
the 30th term is.
So you can do this off to the side.
I am going to include it in the formula and show you all this work at once.
We need to know what that A sub n value is.
We need to know what that A sub 30 term is.
We are going to do the sum of the first 30.
That is going to be equal to n over 2, or 30 divided by 2 times again A1... we know
what that value is... it is equal to 2... now plus A sub n.
I don't know what A sub 30 is, either in your formula or off to the side just go figure
it out.
A sub 30...
Now if you have forgotten.
Arithmetic Sequence, the general form of an arithmetic sequence is...
That is the same color.
Let's...
I am out of colors, so it is going to be green to match this example.
An equals A1 plus n-1 times d.
So I am going to apply this formula inside the second part of my partial sum formula
and find out that A sub n value.
Ok..
A1 is 2 plus n-1, and n is equal to 30 in this example, so 30-1 times my common difference
of... we identified it earlier...
the common difference of 9.
So A1 plus n-1 times my common difference of 9.
1, 2, 3, open parenthesis and 2 open, 3 open, 2 open, 1 open and I am missing a parenthesis.
Then simplify this out.
Well, we have 30 divided by 2 is 15.
2 plus...
We have got 2 plus 30 minus 1 is 29 times 9.
Of course I don't want to struggle with this and take up your time.
So 29 times 9 is 261.
We have 15 times 2 plus (2 plus 261).
That is going to give us 15 times 2 plus 263.
Which of course is just going to be 265, so I should just go ahead write that.
Then 15 times 265 is 3975, or three thousand nine hundred seventy-five.
That is just a couple of examples, pretty straight forward examples of how to find partial
sums of an arithmetic sequence.
One a little bit harder than the other.
Now our third example is again going to be a partial sum, but my problem is written in
summation notation... or sigma notation.
So you know..
What is that partial sum?
Now you might not look at this and immediately go that is arithmetic.
But multiplication is repeated addition.
If I do 4 plus 4 plus 4 which is 12.
Well that is just 3 times 4.
So multiplication does display a short hand version of repeated addition.
This is indeed going to be arithmetic.
If you are not sure.
If you need a more concrete example of that, take the value of 1 and plug it in.
take the value of 2 and plug it in... take the value of 3 and plug it in... and build
out that sequence for yourself.
You will see that you do have a constant addition.
Ok.
So this is arithmetic.
I want the sum of the first 17 terms.
Not because my upper limit or my upper index is 17, but because it is a combination of
the upper and lower index.
If this where say 2 and 17, then we would not be adding 17 values we would be adding
16 terms or values.
So look at your lower index and your upper index and make sure that you pay attention
to how many terms are actually in that sequence.
So if we start counting at 1 and go up to 17... you know... that is what we have.
17 terms.
So..
Let's see.
To find the partial sum of... and it is going to be 17 terms so I can take that n out.
It is going to be n divided by 2 times the first term.
I don't know what the first term is.
We are going to take the k value of 1 and plug it in.
So A sub 1 is going to be 2 times 1 minus 3, or 2 minus 3 is -1.
So A1 is equal to negative one.
My upper limit is 17.
So I need my last term, because it is n/2 times A1 plus An the last term in the sequence.
Let's get this out the way.
Let's plug in 17.
A sub 17 is equal to 2 times 17 minus 3.
That is going to be 2 times 10 is 20, 2 times 7 is 14, so we have 34 minus 3.
That is 31.
So A sub n is 31.
It is the same kind of question.
It is just how it is presented can change the way that you work out the problem, or
at least the way that you think you should work out the problem and make it slightly
more difficult.
17 divided by 2 is going to be 8.5, or we can just leave it as 17/2.
Maybe we will do that.
Here we have -1 plus 31 which is going to be 30.
When you start doing arithmetic with fractions and some of your factors or terms are fractions
and other ones are whole numbers, you want to really just let everything look like a
fraction so you can see how it lines up properly.
When you multiply fractions you do go straight across.
However, if you notice that some stuff will cancel, that some factors will cancel, then
you might want to go ahead and reduce those fractions a little bit.
We are going to cancel across this multiplication sign.
30 divided by 2 is going to cancel out and give you 15 over 1.
17 times 15 is equal to.. where am I at?
There we go!
255.
I am Mr. Tarrou.
BAM!
Go do your homework:)