Integrated math 3
Get a front row seat to Sal's step-by-step solutions to finite geometric series. Learn how to find the sum of the first 50 terms of a series by multiplying each term by a common ratio. Discover how to apply the formula for the sum of a finite geometric series. Plus, explore how to handle series with a change in sign.
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- this thing is sooo confusing(26 votes)
- I know this is an old post, but for the newbies who are still confused this is what I've understood so far:
Basically for the first few minutes he used the formula described in the previous video "Finite geometric series formula". That formula was plugged in for the first few minutes.
In the first problem he uses the formula and then to simplify it further he simplifies the denominator which should be simple enough (1-10/11 = 1/11) and then he multiplies the entire thing by 11 to get rid of the denominator. After doing this he's simplified it enough to get the SUM OF ALL OF THE TERMS. That's what the formula is for.
If you can understand this much, it's easier to carry on with the other problems. If you don't understand, go to the previous videos in the Algebra 2 course - Unit 3 - Lesson 7 (Videos 1 and 2). That's your best bet to understand it.(1 vote)
- for the last problem wouldnt it be to the 31rst power?(12 votes)
- Good question. It's a little complicated, but:
The formula for the sum was made for a sequence of n terms. If there are 30 terms (n=30), and if the 1st term is 10, you multiply by 9/10 from i=2 to i=30 (29 terms - so you have (9/10)^29) plus 1 more for i=1 (the first term, 10) makes 30.
So if he had set it up like the earlier problem "10 + 10(9/10) + 10(9/10)^2 + ... + 10*(9/10)^29" then you add 1 to the 29. But when he just says "there's 30 terms" or "29 terms", you don't add anything.(9 votes)
- if you keep on multiplying 1 by 10/11 for an infinite amount of times, would it eventually become zero? Thanks in advance!(9 votes)
- That is the correct intuition to have. Obviously, we don't reach 0 after any finite number of multiplications, but we say the product tends to 0 or approaches 0.(14 votes)
- At4:55, it says a sub i is equal to a sub i minus i times 9/10. Is this supposed to be i-1?(5 votes)
- It's confusing... the dot is actually the multiplication symbol, not a dot on a "i".
So, Sal does have "i-1"(14 votes)
- Can someone help me understand how to find the first term when given the common ration and sum of the first n terms? I'm having trouble with the solving part. I get to the equation, but then it gets complicated trying to find a with fractions and exponents in the equation. Thanks in advance!(6 votes)
- a=S(1-r)/(1-r^n).(S-sum till n terms, n-number of terms, r-common ratio,a-first term)
If you have to find the first term given the sum, and common ratio, substitute the value of S, r and n and do the calculations carefully. You will get the value of a with this method.(4 votes)
- When Sal simplifies 1(1- -.99^80) why does he simplify this portion to 1-.99^80 instead of 1+.99^80? It's written at4:27(3 votes)
- You may have forgotten to look at the parenthesis. As Sal writes it, it is (1 - (-0.99)^80). Since 80 is an even exponent, both -0.99 and +0.99 will output to the same number, so you can effectively cancel out the second negative sign, giving you 1 - 0.99^8. Since exponentiation happens before addition, we can do this and canceling out the two negative signs would be wrong.(5 votes)
- Wait at6:20, I thought dividing by 1/10 was multiplying by 10 not 100??(4 votes)
- it is, but the numerator is already being multiplied b a 10. so 10*10 = 100. Or shown more explicitly, 10(1-(9/10)^30)/(1/10) = 10 * 10(1-(9/10)^30) = 100(1-(9/10)^30).(3 votes)
- What is the difference between finite/infinite and convergent/divergent geometric series?(2 votes)
- A finite geometric series is a series of the form sum n=0 to k of ar^n. Note that this type of series has finitely many terms, and so the sum always exists.
An infinite geometric series is a series of the form sum n=0 to infinity of ar^n. Note that this type of series has infinitely many terms. For this type of series, the question of convergence vs. divergence arises.
Note that the sum of this type of series is defined as the limit of the sequence of partial sums (that is, sums of finite series of the form sum n=0 to k of ar^n) as the number of terms approaches infinity. If this limit of partial sums exists (which occurs when either a = 0 or |r| < 1), then the infinite geometric series is convergent and the sum equals the value of this limit. If this limit of partial sums does not exist (which occurs when both a is nonzero and |r| >= 1), then the infinite geometric series is divergent and the sum does not exist.
In summary: finite/infinite refers to the number of terms of the series, but convergent/divergent refers to the type of limiting behavior of the sequence of partial (finite) sums for an infinite series.
Have a blessed, wonderful day!(6 votes)
- Why is dividing by 1/11 the same as multiplying the numerator by 11?(1 vote)
- Hello Jyotika,
The answer is that when you divide by a fraction, the number gets larger. For instance, when you divide 1 by 1/4, you get 4, because 1/4 goes into 1, 4 times. Now, when you do the same thing, multiply 1 times 4, you get four. This is because you are doing an inverse operation on an inverse number, essentially doing the same thing with both, and as such, getting the same number. Hopefully this helps you.
Have fun doing your math, and I hope you succeed in your endeavors.(7 votes)
- at2:53, he said 80 but shouldn't it be 78 because of n-1?(3 votes)
- I think you meant 79, but don't be confused:
The reason why he said there are 80 terms is because he included the first term, "1", and then up to the 79th exponent.
n = number of terms
You don't have to worry about n-1.(2 votes)
- [Voiceover] We're asked to find the sum of the first 50 terms of this series, and you might immediately recognize it is a geometric series. When we go from one term to the next, what are we doing? Well, we're multiplying by 10/11, to go from one to 10/11, you multiply by 10 over 11, then you multiply by 10 over 11 again, and we keep doing this, and we wanna find the first 50 terms of it. So we can apply the formula we derived for the sum of a finite geometric series and that tells us that the sum of, let's say in this case the first 50 terms, actually let me do it down here, so the sum of the first 50 terms is going to be equal to the first term, which is one, so it's gonna be one times one minus, let me do that in a different color, one times one minus the common ratio, so the common ratio here is 10/11, 10/11 to the 50th power, to the power of how many terms we have, all of that over one minus our common ratio. And so I'm not gonna solve it completely, but we can simplify this a little bit, this is gonna be one minus, let me put parenthesis here just to make sure we're not just taking 10 to the 50th power. So one minus 10/11 to the 50th power over, this is 11/11 minus 10/11 is one over 11, and so this is the same thing as multiplying the numerator by 11, and so this is gonna be equal to 11 times one minus 10/11 to the 50thpower, and you can try to simplify this even more, but this gets up pretty far, at this point it is just arithmetic. Let's do another one of these, this is kinda fun. So this is more clearly a geometric series, and let's just first think about how many terms we're gonna take the sum of. You might be tempted to say "Okay I'm gonna take it to the 79th power, "there must be 79 terms here", but be very careful, because the first term is when we're taking things to the zeroth power, we're taking 0.99 to the zeroth power. The second term is where we're taking it to the first power, the third term is where we're taking it to the second power, the fourth term is where we're taking it to the third power so on and so forth, so this right over here is the 80th, the 80th term, 80th term. So we wanna find S sub 80, and so this is gonna be equal to our first term is gonna be one times one minus our common ratio to the 80th power, to the 80th power, all over, and I'm leaving a blank because we still need to figure out our common ratio, all over one minus our common ratio. So at first you might say well maybe the common ratio here is 0.99, but notice we have a change in sign here, and the key thing is to say well to go from term to the next what are we multiplying by? Well to go from the first term to the second term, we multiply by negative 0.99. And then, so we're multiplying by negative 0.99. Now to go to the next term, we're again multiplying by negative 0.99, so the common ratio is not positive 0.99, but negative zero, negative 0.99, so let me write that, negative 0.99, and of course that is going to be to the 80th power, all over one minus negative 0.99. And so we could simplify this a little bit, this is all going to be equal to, oh that one we don't have to worry too much about that, and so this is going to be one minus, so negative 0.99 to the 80th power, I should put parenthesis there to make sure we are taking the negative 0.99 to the 80th power. Well, we're taking it to an even power, so it's going to be positive, so that's going to be the same thing as 0.99 to the 80th power, and all of that over, well subtracting a negative that's just gonna be adding the positive, so all of that over 1.99, and we could attempt to simplify it more but, if we had a calculator we could actually find this exact value or close value actually, most calculators don't give you the exact value when you take something to the 80th power, but this is what that sum is going to be. Let's do one more of these. Alright, so here we have a series defined recursively and so it's useful to just think about what it would actually look like. So the first term is 10, and then the next term, so the second term A sub two is equal to A sub one times 9/10, alright. So the next term is gonna be the previous term times 9/10, so it's gonna be 10 times nine over 10, and then the next term is gonna be that times, is gonna be the second term, the third term is the second term times 9/10, so 10 times nine over 10, nine over 10 squared. And the way it's written right now, we don't have it written as a finite geometric series, so let's say we wanna take the sum, let's say we want the sum of first, first I don't know 30 terms, sum of first 30 terms. So what will this be? Well we're gonna take S sub one, S sub 30, oh I wrote ten, S sub 30, the sum of the first 30 terms, is gonna be equal to the first term, we've done this before, the first term times one minus the common ratio, one minus the common ratio to the 30th power, all of that over one minus the common ratio. And let's see we could, one minus 9/10, this is 1/10 right over here, you divide by 1/10, this is the same thing as multiplying by 100, so this is gonna be 100 times one minus 9/10 to the, oh let me write it this way, 9/10 to the 30th power. And, actually these parenthesis you always wanna put parenthesis there to make sure we see we're taking both the nine and the 10 or the 9/10, the whole thing to the 30th power, not just the nine, so there you go, did I.. yep there you go we're done.