Tutor.com Sequences and Series Session

Mar. 24, 2012

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Session Transcript - Math - Algebra II, 3/14/2012 10:27PM - Tutor.comSession Date: 3/14/2012 10:27PM
Length: 26.4 minute(s)
Subject: Math - Algebra II


System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Guest (Customer)
[00:00:00] Problem: Write the next term of the sequence, and write the rule for the nth term. 5,25,125,625,... 4,10,16,22,...

Sarah M (Tutor)
[00:00:08] Welcome to Tutor.com ! Let's get started.

Guest (Customer)
[00:00:14] hello

Sarah M (Tutor)
[00:00:19] I see we are working on number sequences.

Guest (Customer)
[00:00:18] thanks
[00:00:41] yes i got the next term

Sarah M (Tutor)
[00:00:50] Ok, what did you find for the next term?

Guest (Customer)
[00:01:01] im just stuck on the rule
[00:01:18] i got 3 for the first sequence
[00:01:32] and 28 for the second

Sarah M (Tutor)
[00:01:54] For the first list you would have 5, 25, 125, 625, 3 ?

Guest (Customer)
[00:02:09] hold on a sec

Sarah M (Tutor)
[00:03:16] Let me know when you are ready to continue.

Guest (Customer)
[00:03:40] sorry , my bad 3125

Sarah M (Tutor)
[00:03:55] I agree with that. We need to find the pattern. What did you do to get that number?

Guest (Customer)
[00:04:22] i multiplied by 5

Sarah M (Tutor)
[00:04:46] Good, for each term we multiply by another 5. Let me write it out for a few of them and we can find an easier way to write the general form.

Guest (Customer)
[00:04:55] ok

Sarah M (Tutor)
[00:05:22] Can you think of a different operation that is a shortcut for multiplying a number by itself over and over?
[00:06:06] For example, if we want 125, what would we do to the number 5 to get there in one step?

Guest (Customer)
[00:06:44] im not sure

Sarah M (Tutor)
[00:07:16] That's ok. We can use an exponent. So, 25 is 5^2 and 125 is 5^3. Exponents just say, "multiply that by itself a few times."
[00:08:08] To write the rule we want to generalize this. For the first term our exponent is 1. For the 2nd term our exponent is 2. What might we use as the exponent on our "nth" term?

Guest (Customer)
[00:08:24] will the rule be a1* 5^r-1
[00:08:28] ??

Sarah M (Tutor)
[00:08:47] It will be much simpler than that. For the 5th term we had 5^5. Can you guess what exponent we'd use for the 7th term?

Guest (Customer)
[00:09:18] 6

Sarah M (Tutor)
[00:09:50] Let's look at our pattern. For the very first term our exponent is 1. For our 2nd term the exponent is 2. For each spot in our pattern we use another 5.
[00:10:03] So, for spot 6 we'd take 5^6. For spot 7 we'd take 5^7.
[00:10:25] To generalize we just need to use a variable for some term way out there down the lin.
[00:10:27] line.

Guest (Customer)
[00:10:46] oh ok gotcha

Sarah M (Tutor)
[00:11:05] So for that term "r" way down the line we'd have to multiply 5*5*5*5... "r" times. Right?
[00:11:47] Do you have any questions on why that rule fits our pattern?

Guest (Customer)
[00:12:40] ?um so the rule is just 5^r

Sarah M (Tutor)
[00:13:11] Right. We can test it... using the method of multiplying a 5 times the previous number, what would be the next term in the list?
[00:14:28] 3125*5 is 15625, right? We can see if that agrees with our rule, 5^6. What do you think?

Guest (Customer)
[00:14:47] yup i agree, that what i got

Sarah M (Tutor)
[00:15:01] Great! Do you have any questions on how we did this one before looking at the next list?

Guest (Customer)
[00:15:36] no got it

Sarah M (Tutor)
[00:15:48] Ok, and you said you had the next term for the next list, right? What value did you find?

Guest (Customer)
[00:16:16] 28

Sarah M (Tutor)
[00:16:27] Good! How did you find that value?

Guest (Customer)
[00:16:42] add six

Sarah M (Tutor)
[00:17:05] Great. I'll write out a few of the steps and then we'll generalize it.
[00:17:55] Actually, before I do that. Are you working with arithmetic and geometric sequences in class yet?

Guest (Customer)
[00:18:33] yes i am

Sarah M (Tutor)
[00:18:52] Have you seen the general formula for an arithmetic sequence?

Guest (Customer)
[00:19:48] yes i have

Sarah M (Tutor)
[00:20:16] That is basically what we need here. An arithmetic sequence is when you add a constant to the previous term.
[00:20:23] I'll write that form down and we can discuss it.

Guest (Customer)
[00:20:29] ok

Sarah M (Tutor)
[00:21:13] Ok, so let's take this piece by piece. a1 is the first term. What is our first term?

Guest (Customer)
[00:21:45] 4

Sarah M (Tutor)
[00:22:04] Good. The (n-1) stays because we want this to be for some generic value "n".
[00:22:30] d is the "common difference." It's the constant that we add to each term. What did you find that to be?

Guest (Customer)
[00:22:58] 6

Sarah M (Tutor)
[00:23:44] That is the rule for this list. Instead of adding 6 over and over, we always start with 4 and then go from there.
[00:24:00] So, if we wanted to get that 16, these are the two ways we could do it... one way by adding 6 and the other using our rule.
[00:24:33] 16 is the 3rd term, so (3-1).
[00:25:07] In both cases we add 2 sixes.
[00:25:11] Do you have any questions about this rule?

Guest (Customer)
[00:25:28] nope got it

Sarah M (Tutor)
[00:25:42] Great. Do you have any questions on these two rules before we end the session?

Guest (Customer)
[00:26:06] no thank you so very much!!

Sarah M (Tutor)
[00:26:14] Thanks for using Tutor.com. Please fill out the survey as you leave so we can learn how we can better help you in the future! 
[00:26:17] Have a good evening!