Main content

## Precalculus

### Course: Precalculus > Unit 9

Lesson 2: Geometric series (with summation notation)- Summation notation
- Summation notation intro
- Geometric series with sigma notation
- Worked example: finite geometric series (sigma notation)
- Finite geometric series
- Finite geometric series word problem: social media
- Finite geometric series word problem: mortgage

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# Finite geometric series word problem: social media

Watch Sal solve an example of using a geometric series to answer a fun word problem. Created by Sal Khan.

## Want to join the conversation?

- Sorry, but can anyone explain to me how Sal arrived at 1.47? Is " 1 " used to represent a month?(18 votes)
- Let say 1st month user = 50; and we need to add 47% each month.

To get 2nd month we need to have add first month user + 47% from 50 or 50 + (50 * .47)

We know 50 * 1 = 50 and 50 * .47 = 23.5

Combined together to get 2nd month we get: 50 * 1 + 50 * .47 = 73.5

Then factoring: 50 * (1 + .47) = 73.5

Or 50 * (1.47) = 73.5

Conclusion, we can get 2nd month by 50 * 1.47(8 votes)

- I think there must be a difference in the use of the word "through" between US and Australian English. I understood "through month n" to mean "from the start of month n to the end of month n" but from the math I gather that it means "from the start of the year to the end of month n". Could someone please confirm or deny that this is how you use "through"?(29 votes)
- I understand what you are asking.

I believe that the issue is interpretation of an accidentally unclear problem more than linguistic barriers. I can now see it from both perspectives, but I agree that in this problem - "through" means how many users were added from the beginning of the year to the end of month n.(15 votes)

- Why not the first answer, 50(1.47^n)?

- I see, by “new” users, the question is looking for how many users were added that month, not total users. I misunderstood the question.(14 votes)- For new users added each month though, couldn't you just do 50*1.47^n-50, aka final - initial?(5 votes)

- Where might I find more solved examples of finite geometric series word problems? I am having great difficulties solving and visualizing them, and the two videos here are clearly not enough for me.(6 votes)
- Here are links to a couple of sites with word problems involving series. They have a mix of arithmetic and geometric.

1) http://www.mathbitsnotebook.com/Algebra1/Functions/FNSequencesWordPractice.html

2) http://regentsprep.org/Regents/math/algtrig/ATP2/SequenceWordpractice.htm

You may be able to find others. I found these doing a search on "geometric series word problems"(3 votes)

- How did Sal get 50 x 1.47^n at the end? How does 50 x 1.47^n-1 + .47 = 50 x 1.47^n(2 votes)
- Well, you start the month with a certain number, then you add some during the month, and the last column is the total number at the end of each month.

On the nth row, we start with 50 times (1.47)^(n-1)

then we add

50 times (1.47)^(n-1) times 0.47

So by the end of the nth month we have:

50 ∙`(1.47)^(n-1)`

+ 50 ∙`(1.47)^(n-1)`

∙ 0.47

To simplify, we can factor out a 50 to leave

50 [`(1.47)^(n-1)`

+`(1.47)^(n-1)`

∙ 0.47]

now if we factor out the remaining common factor of`(1.47)^(n-1)`

we get

50 (`(1.47)^(n-1)`

) [1 +∙ 0.47]

That 1 +0.47 is just another 1.47, so we now have

50 (`(1.47)^(n-1)`

) (1.47)

50`(1.47)^(n-1+1)`

50`(1.47)^n`

(8 votes)

- this video is different from the other problems and what did the third sentence mean by n is grater than or equal to 1 and n is smaller than or equal to 12?(1 vote)
- It's just saying that the expression will only be valid if n is some value from 1 to 12. The social media site only made a claim about one year, and there are 12 months in a year!(5 votes)

- Couldn't it be expressed also as ∑ where n=1 to 12 with 50,000(0.47)^(n-1) ?(2 votes)
- Not quite. It would be ∑ where n=1 to 12 of (0.47)(50)(1.47)^n-1.

If you look at the right answer choice, the constant coefficient is (0.47)(50) while only (1.47) is being taken to different powers.

I hope this helps you understand the answer in terms of a summation better!(4 votes)

- If suppose I get this question in an exam then how would I know it forms AP or GP by reading the problem?

I am sorry but I am unable to distinguish despite knowing that AP is what we get by add/subtract and GP by multiplication

Please help(1 vote)- Think about what happens to the initial term, in this case the 50,000 social media users.

In February they will have increased by 47%, which would be an increase by 23,500 users to 73,500.

If this is an arithmetic sequence, then the users will increase by 23,500 every month, but 23,500 is not 47% of 73,500, so it can not be an arithmetic sequence.

Rather, the number of users would increase by 34,545 to 108,045.

If this is a geometric sequence, then we should see that the ratio between two consecutive terms is constant, and sure enough

108,045∕73,500 = 73,500∕50,000 = 1.47, which makes sense since we're dealing with a 47% increase every month.(4 votes)

- Is there a simple formula to do these kinds of problem in 2 minutes? For Sat MATH 2(2 votes)
- How did Sal get the first term, 50? I thought the first term would be 50,000(1 vote)
- In the question, it wants you the number of new users in thousands instead of normally. Since 50,000 / 1,000 = 50, 50 is the correct first term.(2 votes)

## Video transcript

A new social media site
boasts that its user base has increased 47% each
month for the past year. The number of users on January
1st of last year was 50,000. Which expression below
gives the total number of new users in thousands
that were added through month n of the past year, where 1 is
less than or equal to n, which is less than or equal to 12. And they give us some
choices of expressions for the total
number of new users that were added through month n. And I encourage you to
now pause this video and try to think about
which of these expressions actually show that,
that gives that value. Well to tackle it, I'm going
to make a little bit of a table here. So let's say we have month, and
then we have starting users-- so users at the
start-- and then let's think about the users added. I want to give myself
some space to work with. And then users at
the end of the month. So in month 1, which is
January we can assume, we started with 50,000 users. They want us to write
the expression in 1,000. So we started with 50,000 users. And how many did we add? Well we added 47% of 50,000. So 50 times 47%. So times 0.47. So how many do we end with? Well 50 plus 50
times 0.47, that's going to be 50 times-- I'm gonna
do the 50 in green-- that's going to be 50 times 1.47. If this isn't clear,
just think about this. This you could
rewrite as 50 times 1. So 50 times 1 plus
50 times 0.47, that's going to be
50 times 1 plus 0.47. Or 1.47. So it's going to be this
thing right over here. So now let's go to month 2. We start with what we
ended the last month. So I could just copy
and paste this actually. Let me just do that. So copy and paste. So that's what we start with. Now what are we going to add? Well we're going to add
this, what we started with, times 0.47. And so what are we
going to end with? Well if you sum these two-- and
you could write it this way-- this is going to be
this thing times 1.47. Or we could just write this
is 50 times 1.47 squared. And you might start
seeing a pattern now. Let's go to month 3. So month 3: what do we
start the month with? We start the month
with this thing. Let me copy and paste this. So, copy and paste. We start with that. What do we add? Well we're going to
take that, and we're going to multiply it times 47%. We're going to
multiply it times 0.47. And so what are we
going to end up with? We're gonna have this times
1 plus this times 0.47. That is going to be equal
to that times 1.471. Or we could just write this as
50 times 1.47 to the 3rd power. So what's the pattern here? Well in each month, we're
going to be starting with 50 times 1.47 to a
power 1 less than the month. In the third month,
the power here is 2. In the second month,
the power here is 1. In the first month, the
power here-- you don't see it but you can view this as
times 1.47 to the 0 power. So 1st month, 0 power. Third month, you
have the 1st power. Third month, you have the
second power over here. So if we're thinking
about the nth month this is going to
be 50 times 1.47 to the n minus 1th power, is what we're
going to start the month with. Now what are we going
to add in the nth month? Well it's going to
be that times 47%. So it's going to be--
we'll just copy and paste that-- so it's going
to be that times 47%. Times 0.47. And then, what are
we going to end with? Well when you add
these two things, you are going to get
50-- I'll just do it in the right colors instead
of copying and pasting it-- you're going to get 50
times 1.47 to the nth power. So let's think about
how we can come up with the expression for the
total number of new users in thousands that were
added through month n. So there's a couple of
ways to think about it. You could say, well how
many total new users did we have at the
end of month n? Well at the end of month
n, we had that many. And then how many did we have
at the beginning of the year? Well we have 50,000. So how many total new users
did we add through month n. So we finished with this much. Let me just write it down. So we just finished
with that much. And let me paste that. So that's what we finished with. And we started
with 50,000 users. So this is essentially how
many we added through month n. Now do any of our
expressions look like that? Well, no not quite. If this one had a minus
50 right over here, if that said minus 50, then
that would've done the trick, but this doesn't do it. And none of the others really
seem to either have this form or seem to be
something that would be very easy to
manipulate in this form. So that's one way to do it, but
that's not one of our choices. So what's another way
of thinking about it. Well, we could literally
just add how many new users we added month by month. So we literally could just
add all of these things right over here. So let's see, we could literally
just add all of these terms. But let me simplify
it a little bit. So what are some common factors
that we see in all of these? Well we see they all have a 50. And they all have something
being multiplied by 0.47. So let's factor out
a 0.47 and a 50. So let's factor that out. So this is going to
be equal to-- if we were to sum all this up-- it's
going to be 0.47 times 50. And then what's left over? Times-- so in the first
month, if you factor those two things out, you're going
to just be left with a 1. In the second month if you
factor out the 50 and the 0.47, you're left with 1.47. In the third month, if you
factor that and that out, you're left with 1.47 squared. And we're going to go all
the way to the nth month. If you factor out
that and that, you're left with 1.47 to the
n minus 1th power. So which of those
expressions look like that? Well this is exactly the second
expression right over here. This is exactly what
we came up with. And we're done.