Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. 1 * 5 * 5 * 5 * 5 * 5 S Our a sub N term is going to be whatever the first term was our a sub one times our ratio to our n - 1 power. If we look here at our fifth term, the exponent was four. On our fourth term the exponent was 3. So our generic term for our a sub N is going to be a sub 1 * r to the n - 1. So if we have a sub one equaling 5 and a sub one or and R equaling 3, it wants us to come up with our first few terms. So 5 * 315 our a sub three would be 5 * 3 to the 3 - 1 or 5 * 3 ^2 5 * 945 our a sub four 5 * 3 to the 4 - 1 five times 3 ^3 5 * 27 is 135. If we look at the next example, our A sub 2 is going to be -5 times A to the 2 - 1 or -5 * A to the 1st or -5 * -6 positive 30. Our A sub three would be -5 A to the 3 - 1 or -5 * a sub two. Our a sub two had been 30. So we get -150 our a sub 4 -5 A 4 -, 1 negative 5A, sub three negative 5 * -150 would be +750. If we look at these next couple, we want to figure out a sub twelve. Well, it's going to be our first term, our common ratio to our n -, 1. So 5 * -2 to the 11th power 5 * -2 to the 11th is going to be -10,240. If I wanted a sub 40, my first term times negative 1/2 to my 40 -. 1 power 1000 * -1 half to the 39 and I'm going to put that into my calculator and I'm going to get -125 over 68719476736. Wow. OK the next type of problem they give us a listing and they want us to figure out the formula for our A sub N and then they want us to find the A sub seventh term. So our A sub N is going to be the first term. Our ratio here would we multiply 3 to get 12, so our R would be 12 / 3 or 4. So 3 * 4 to the north -1 if we wanted A to the 7th, we'd have 3 * 4 to the 7 - 1 three times 4 to the 6th which is 12,288. This next one our a sub N our first term .0004, our ratio -.004 / .0004. So we're going to get -10 to our n - 1. So if we wanted A to the 7th .0004, so 10410 thousandths times -10 to the 6th. So once again, I'm going to just put this into my calculator and come up with 400. Now we want to come up with what happens if we're going to do a summation of all these terms. So our S of N term would be a sub 1 + a sub 1R plus a sub one r ^2 + a sub one to the RN -2 + a sub one R to the n -, 1. Now, if I thought about taking this equation, I'm going to now multiply that equation through by an R term. So what that really does is it makes each of those terms have one more R in its exponent. Let's just go... dot for a second, A1R to the north -1 and then AR to the north. If I go through and take that original top equation and multiply everything through by R, it just makes a one turn into A1R and A1R turn into a one R-squared and A1 R-squared would turn into A1R cubed all the way down to that last term would turn into ARN. Now I'm going to subtract those two equations. When I subtract the two equations I get SN minus RSN equaling a sub one. The A sub one Rs are going to cancel the A sub one R twos are going to cancel all the way down to leaving us that A sub RN I'm going to factor out an S sub N on the left side. I'm going to factor out an A sub one on the right side. There should have been A1 back here. And then if I want as sub N by itself, I get a sub 1 / 1 -, r to the n / 1 -, r So that's our summation formula for a geometric series. Now we want to talk about what happens if R. Let's say if the absolute value of R is less than one. Or another way of saying that is, if R is in between -1 and one. What happens to R to some power if N is really, really big, or if N is approaching Infinity? Well, if we thought about Infinity, we thought about using 1/2, maybe what's 1/2 to the 10th and then 1/2 to the 100th and then 1/2 to the thousandth. That's getting closer and closer and closer to 0. So if it was an infinite series that we were trying to find the sum of, we would have our S equaling our a sub one 1 -, r to the north. But that R to the north would be 0 if the R absolute value of R is less than one. So that's going to equal SA 1 / 1 -, r when our ratio absolute value is less than one. So the sum of the first N terms of a geometric series is just a formula S sub N equal A to the one times the quantity 1 -, r to the north over 1 - r, where R is not equal to 1. And the sum of an infinite geometric series is a sub 1 + a sub one r + a sub one r ^2 + a sub one r ^3 plus... S equal a sub 1 / 1 -, r when the absolute value of R was less than one. If the absolute var is greater than or equal to 1, the infinite series doesn't have a sum because we're always adding a bigger number and adding a bigger number. The sum of the 1st 12 terms then. So if we want the sum of the 1st 12 terms, we're going to have our S sub N or our S sub 12 is going to equal to 1 minus our ratio here 6 / 2, which is 3. So 3 to the 12th power over 1 - 3, so 21 minus whatever 3 to the 12th is 3 to the 12th. 531441 all over -2 So we're going to get 531440 as our final answer there. If we wanted the 12th terms here, we'd have -3 halves 1 minus our ratio. Here 3 / -3 halves 3 / -3 halves means 3 * -2 thirds. So we'd get -2 as our ratio to the 12 / 1 - -2. Grabbing our calculators and plugging this all in, we would get 4095 over 2. Looking at this next one, we want to find the indicated sum. So we're going to start by remembering that that 5 is a constant and it can just come in and out. So if we have two to the I, if we thought about this, our first term is going to be two, our second term would be 4. So our ratio is going to be 4 / 2 or two. So we're going to have five times our A sub one, which would be two, our one minus our ratio to the 10th term all over our one minus our R. So if we plug this into our calculator, five times the quantity, 2 times the quantity 1 -, 2 to the 10th. And then we're going to take all of that and divide it by a -1, so 10230 for the next type. They want us to find the sum of the infinite geometric series, so 3/4 / 3 for my ratio. So my ratio here is going to be 1/4, and that is having the absolute value less than one. So our S is just going to be our first term over 1 minus RR. So 3 / 3/4 is just going to be 3 / 3 fourths, 3 * 4 thirds, so a sum of four. This next one, our ratio negative 1/2 / 1, so negative 1/2. Our sum is our first term over 1 minus our ratio, so 1 / 1 -, a negative 1/2, which would give me 3 halves, so our sum would just be 2/3. This next type they want us to express each repeating decimal as a fraction in lowest terms. So our ratio is going to be 51051 hundredths divided by 5/10. So if we simplify that, we get 110th. So our sum is going to be the first term over 1 minus our ratio. Now the easiest way to simplify that is to multiply by 10 / 10. So we'd get 5 / 10 - 1 or five ninths. The next one, our ratio is going to be 47 / 100 ^2 / 47 / 100. So we're going to get 1 / 100. So this fraction is going to be the 47 / 100 / 1 - 1 one hundredth. Once again, the easiest way to simplify these is just multiply by 100 / 100 this time. So 47 / 99 is the same thing as that .47 repeating. Thank you and have a wonderful day.