Log in Hunter Priniski 10 years agoPosted 10 years ago. Direct link to Hunter Priniski's post “I came up with the formul...” I came up with the formula (-1)^(n-1) * 8(n-1)/(2n+1). Is this correct as well? • (2 votes) Daniel Schneider 10 years agoPosted 10 years ago. Direct link to Daniel Schneider's post “Not quite. Your formula ...” Not quite. Your formula fails once we reach a_4, where Sal has a term of -32/9, but your function would give (-1)^(4-1) * 8(4-1)/(2*4 + 1) = -24/9. Yours would work if the numerator were simply increasing in magnitude by 8 each time, but here's it actually doubling. Hope that helps! (8 votes) Momina Hussain 8 years agoPosted 8 years ago. Direct link to Momina Hussain's post “Hey, so I just figured ou...” Hey, so I just figured out that the common ratio between these successive terms is -10/7, therefore shouldn't the fourth term be -160/49 instead of -32/9? • (1 vote) Travis Bartholome 8 years agoPosted 8 years ago. Direct link to Travis Bartholome's post “Well... In a word, no. Th...” Well... In a word, no. The ratio between the first two terms may be -10/7, but the fourth term is part of the series as Sal has defined it, too, so we can't just discount it or say that it must be wrong when we're trying to figure out the pattern. This is not a geometric series, but if you just look at the first two terms, you might think it is. In fact, if you just look at the first two terms of any series, you could convince yourself that it's geometric because there will always be some constant ratio between two given numbers. If we knew that this series was geometric, your statement would be correct; however, we don't know that, and in fact it's not a geometric series, so we can't assume that there will be a common ratio. Hope that makes sense. (7 votes) Sangjin Oh 8 years agoPosted 8 years ago. Direct link to Sangjin Oh's post “what does a sub n mean?” what does a sub n mean? • (2 votes) Dimas Gomez 8 years agoPosted 8 years ago. Direct link to Dimas Gomez's post “a sub n, which people oft...” a sub n, which people often format as a_n is the last number in a series, or any particular number represented by its position n, in the a series (a_1, a_2, a_3, ..., a_n). If we have 10 numbers in some series, the tenth number, a_10, will be the a_n. In the example in this video, the first term is (1 vote) Sam Weaver 9 years agoPosted 9 years ago. Direct link to Sam Weaver's post “How can I identify patter...” How can I identify patterns in series easier? I know it's broad, but that's my main problem when completing these questions. Once I identify the pattern, I can come up with the general term pretty easy. • (2 votes) Stefen 9 years agoPosted 9 years ago. Direct link to Stefen's post “Look for the usual suspec...” Look for the usual suspects: repetition, distance between terms, ratio between terms, alternation patterns. The best method is just to practice it. Look for other resources online, calculus texts, etc. Remember way back when factoring a quadratic seemed difficult, but now, with lots of experience under your belt, no problems, right? Same deal here. (1 vote) Irene Mutuc 8 years agoPosted 8 years ago. Direct link to Irene Mutuc's post “i have a different formul...” i have a different formula! is this right [(-2)^n * 2] / -(2n +1) • (2 votes) kinosh*ta.hideki 10 years agoPosted 10 years ago. Direct link to kinosh*ta.hideki's post “Can i write it as 4*(-2)^...” Can i write it as 4*(-2)^n / 3+2n instead? • (1 vote) Brandon Payne 10 years agoPosted 10 years ago. Direct link to Brandon Payne's post “There are an infinite num...” There are an infinite number of ways to express most sequences/series explicitly. Your expression is close; however, at n = 1 your expression = -8/5. If you look at Sal's expression you will notice his series starts off with n = 2. Thus, if you were to decrease the index of your expression by one you would get a correct answer as well. For example, Notice, if you manipulated the final expression above enough, you would get same answer as Sal. (Focusing only on the numerator) (2 votes) paupaurosario 8 years agoPosted 8 years ago. Direct link to paupaurosario's post “Could another Sigma notat...” Could another Sigma notation for this same problem be: (-5)^n/3n? • (1 vote) Travis Bartholome 8 years agoPosted 8 years ago. Direct link to Travis Bartholome's post “Yes, that would work; I g...” Yes, that would work; I guess you couldn't see the video comments, but that idea was at the top. And actually, you could even go so far as to express it as (-5/3)^n, which I think may be even more useful because it really shows the geometric nature of this series. As another user said, I suspect Sal used the (-1)^n notation because he wanted to use the concept of an oscillator, which is a good thing to be able to recognize and apply. (2 votes) Ryan Hoang 8 years agoPosted 8 years ago. Direct link to Ryan Hoang's post “I know that this sequence...” I know that this sequence may not be completely simplified, but, for sigma terms, is it okay if i do 4*(-2n^n-1)/3+2(n-1)? • (1 vote) ArminBay 8 years agoPosted 8 years ago. Direct link to ArminBay's post “Hey, So could this series...” Hey, So could this series be explicitly defined with another formula? Because I came up with another formula, different from yours, and mine works too. • (1 vote) Ron Jensen 10 years agoPosted 10 years ago. Direct link to Ron Jensen's post “In the previous video (Wr...” In the previous video (Writing a series in sigma notation) Sal takes great pains to use (-1)^n for alternating the sign of the terms, now 5 minutes later in video time he's completely ignoring that concept. Why did he not say (-1)^(n-1) *2^(n+1)/(2n+1) to be consistent? • (1 vote) Ted Fischer 10 years agoPosted 10 years ago. Direct link to Ted Fischer's post “It can be done either way...” It can be done either way. Separating out the (-1)^n term is akin to writing 7 - 6 as (1 vote)Want to join the conversation?
a_2. And a_n would be an infinite unkwown number. Bytheway, there are sums that the infinite sum of n numbers is known, like sum of 2^-n, n=0 to infinity. http://www.wolframalpha.com/input/?i=sum+of+2%5E-n,+n%3D0+to+infinity
4*(-2)^(n-1) / 3+2(n-1)
4*(-2)^(n-1) / 3+2n-2
4*(-2)^(n-1) / 2n+ 1
4 = 2*2
(-2)^(n-1) = (-2)^n / -2
2*2*(-2)^n / -2
2 and -2 cancel out to equal -1
-1*2*(-2)^n
-1 * 2 = -2
-2*(-2)^n = (-2)^(n+1)
Thus, your expression = (-2)^(n+1) / 2n+1
My reasoning is, (-5)^1 = -5, ^2 = 25, ^3 = -125, etc, and it would be much simpler. Would that work as a solution for this as well? Thanks! Sorry, I'm watching the video "Writing a series in Sigma notation" but the comments are from a different video. Not sure what's going on with the site, but I went back and checked other videos and the same thing is happening there. I'll try to put the comment on the right video.
7 + -6. Once you get comfortable with the notation, either is easily recognizable. Wrapping the negative into the exponential base gives you a simpler expression, though.
Video transcript
Let's say that we're told thatthis sum right over here, where our index starts at 2 and wego all the way to infinity, that this infiniteseries is negative 8/5 plus 16/7 minus 32/9plus-- and we just keep going on and on forever. And so what I want to dois to explicitly define what a sub n is here. So right now we justsay, hey, if you take the sum of a sub nfrom n equals 2 to infinity, it turns out you getthis sum right over here. But let's think about what asub n-- how we can actually define it in terms of n. And I encourage you to pausethe video right now and try it on your own. So the first thingthat you might realize is, well, this is the numberthat we're going to get. Let me write it this way. a sub 2 is equalto negative 8/5. a sub 3 is equal to 16/7. a sub 4 is equalto negative 32/9. And I'm just giving the signto the number in the numerator. Negative 8/5 is the samething as negative 8 over 5. Let me make that alittle bit clearer. So I'll make that alittle bit clearer. So this is negative 8/5. Obviously, thisis positive, so I don't have to reallyworry about it too much. And then here, I'm justsaying negative 32/9, so it's the same thingas negative 32 over 9. So let's see ifwe can first find a pattern in the numerator. So when we go from negative8 to 16, what's happening? Well, we're multiplyingby negative 2. Now, to go from16 to negative 32, we're multiplyingby negative 2 again. So you might say,OK, well, whatever we have in the numerator mustbe a power of negative 2. And, all right, if yousay, well, maybe this is negative 2 squared, well,you know that negative 8 isn't negative 2 squared. Negative 2 squared isequal to positive 4. Negative 8-- thisright over here. Negative 8, that is equal tonegative 2 to the third power. 16 is equal to negative2 to the fourth power. Negative 32 is equal tonegative 2 to the fifth power. So notice, our exponenton the negative 2 is always going to beone more than our index. Our index is 2,our exponent is 3. Our index is 3,our exponent is 4. Our index is 4,our exponent is 5. So that gives a sense that atleast the numerator is going to be-- whatever our indexis, it's going to be-- so let me write this down. So a sub n is equalto-- well, it's going to be negative 2 towhatever index we're at, to that index plus 1 power. So that's a reasonable wayto think about our numerator. Now let's think aboutour denominators. So over-- So we go from 5,so when n is 2, we're at 5. When n is 3, we're at 7. When n is 4, we're at 9. So notice, 5 is2 times 2 plus 1. This right over hereis 2 times 3 plus 1. This right over hereis 2 times 4 plus 1. And you should justkind of play around with differentpatterns in your head until you say, hey,well, look, this is increasing by 2 every time. Notice, this increasesby 2 every time. But these aren'texactly multiples of 2. These seem to be off by one morethan the multiples of 2, which is a good signthat this is going to be 2 times our index plus 1. So we could write this as2 times our index plus 1. And we're done. That's what a sub n is. And if we wanted to writethis series in sigma notation, we would write thisas the sum from n equals 2 to infinityof negative 2 to the n plus 1power over 2n plus 1. And that would equal thisseries right over here.
