Formula for a finite geometric series | Sequences, series and induction | Precalculus | Khan Academy
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- Опубліковано 7 вер 2024
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Thank you, Sal, for creating Khan Academy. I'm very grateful that you share your expertise, for Free. (Private math tutors are expensive...and besides that, you explain it better than any instructor I've ever heard)
great explanation and I understand the proof, but I'm amazed how someone came up with it, looking at the beginning where would they get the intuition to multiply the sum by r and subtract from the sum
This baffles my mind as well, did you get the answer in the meantime?
Thanks for sharing. Normally we take S sub n as a sum of first n terms and you write here it as the sum of first n+1 terms. It really didn't disturb, but to make things easier, I just mentioned it.
this video is missed in the series(Precalculas) course. Hopping that it will be added!!!!!!!
is this different from "sum of all terms in geo. prog." Sn= a(1-r^n)/(1-r) ? I am confused. you used exponent of r as n+1 and not n.
Ikrrr this is not the correct formula
0:37 it should be up to n-1 not n right? since n is the number of terms in the series and you need to subtract 1 because you're using ar^0?
why so we even go to school is it that simple? waw I'm amazed thank you so much
LOVE THIS!
Again you are an awesome teacher. I kind of understand this because I am only in sixth grade.
melaina Math genius u may think this is pretty complicated since uve mentioned that u "kind of get this" but it really isn't and its a really good thing that ur already taking an interest in higher level math. Look into the actuarial science career or financial mathematics if u don't already know what they are. Plus if you start taking preliminary exams for actuarial science early on in high school (as many students do) then trust me, it'll be extremely beneficial for u later on in college. I know you didn't ask for advices or anything but this is something that i really wish someone told me when i was your age.
You took this is 6th grade? I took this in 4th grade 7 years ago. It was pretty easy though. Now looking back it feels like a walk in the park. Nothing is challenging anymore :( I'm bored
Why have ar^n when it became ar^n+1 after multiplying by r in step 2 .. if that would have been the case then in 3rd step when Sn- rSn was being derived we would have a+ar^n -ar^n+1. Please advise
You make this easier to understand than the formula itself, thanks much!
9:00 not really to hard to figure out in your head a fraction to such a high power is a decimal with a lot of 0 so you can practically ignore that term (I doubt the calculator has such accuracy too). And a fraction a/(1/b) is a basic fraction rule: a*b/1, so 2 * 3 = 6.
Thank you
Thx Mr Khan
The formula of the sum of geometric progression is a-ar^n / 1-r . Not the power of n+1
How great are you!
This formula seems different to what I have seen in other videos: (a(1-r^n))/1-r
What's the difference?
This video starts the sequence with the 0th term. So it defines Un = ar^n.
If you change "ar^k" with "ar^(k-1)" and redo the steps shown in the video, it will yield the same result as what you mentioned.
Really late but hope this helps
Forgot to mention that you need to change the lower limit of the sigma notation to 1 instead of 0.
why doesn't subtract the formula further to Sn=ar^n by cancelling (1-r)?
I'm pretty sure that it should be a(1-r^n)/(1-r) and not a(1-r^(n+1))/(1-r)
Actually 'n' in the formula a(1-r^n)/(1-r) denotes the *number of terms* in the series. Sal's series starts from r^0 to r^n. So, the *number of terms* is n+1.
This makes sense tho, my professor derived this formula previously.
Another question is why did you multiply that R x Sn ???
Like why is it we can do that.
I'm no mathematician but I believe people do that to make an equation for concepts like the Geometric Series. I'm pretty sure they don't instantly come up with it how ever. I'm sure they tried multiplying by other numbers or played around with the variables. I just watched this lesson and I'm baffled by how such easy rules can lead to simple yet important equations that we can trust. Hopefully I made you understand better! :)
Hey, can you tell me, why you could shift the rSn by one?
Isn't this also true?
Sn*r +a =Sn +[ar^(n+1)-a]/1-r
As Sn*r is Missing a and Sn is missing ar^(n+1)
The point is I tried to solve this way and got Sn= [ar^(n+1) - a]/1-r which is not same as the one derived in the video.
Can someone explain? :)
try your formula on a geometric serie and compare the results. In my opinion it is possible to derive a different formula if you consider for example that : Sn = a*(r^0)+a*(r^1)+a*(r^1)+........a*(r^-(n-1)); the difference here is that n is the number of terms, whereas the number of terms in Sn formula of the video is (n-1) not n.
The question here is; why multiply r times Sn. Whats the purpose of doing it?
Sergio Prada because that allows us to subtract and thus get rid of most of the terms
How is this precalculus? We didn't even mention infinite series in any of my precalculus courses let alone summation notation. Helpful video though thanks!
His voice boutta make me act up
@2:43 why was r multiplied to Sn? Also, why was rSn subtracted to Sn?
why do we multiply both sides by r
Love having this resource
Can we start from k=1 to k=n?
I though the sigma symbol was only used in a series because it means "the sum of" not sequences because they are just points.
ye, this video is all about series, but as he said in another video he confused it,
Can you help me with my Sequences and Series Exam
pretty close to 6...pretty close to 6
I have a question. What if the function is ∑m=(3m+7)?
That'd be an arithmetic series, which makes this formula invalid to it!
wait im still confused...
isnt series is the sum of sequence??
then why the answer is 6 if you have to add all the 100th terms??
pls help
It adds up to 6 because those 100 terms are tiny, tiny, fractional amounts that get smaller and smaller because (1/2^)n is a smaller and smaller fraction as n increases. (1/2 times 1/2 is 1/4, etc....)
Multiply that tiny (1/2)^n fraction by 3 and you still very small amounts that get smaller as n increases.
This is how you get 6.
Jorge Gutierrez this formula is wrong anyway
Anas Hakouz lmao yea it is it’s supposed to be
Sn=a(r^n-1)/r-1 assuming ratio is greater than 1
Or
Sn=a(1-r^n)/1-r
❤️
How did u get from Sn-rSn to Sn(1-r)?
+Kevin Wang Think of Sn-rSn as Sn*1 - Sn*r
by reversing the distributive property you can factor out the Sn thus getting Sn(1 - r)
Ex:
5*6 - 5*4
=5(6 - 4)
-In both cases you'll get 10
Jamie Brown yeah, I was thinking about all the different kinds of complicated things, and turned out to be distributive property...now i know y my math teacher used to ask "am i over complicating it or under complicating it?" when he didn't understand our question XD
He took Sn out as a common factor.
I think there is a typing mistake: in the geometric sum, the exponent of r should be n & not n+1
Dude, he took r^n in the 1st equation and therefore r^(n+1) in the 2nd whereas it should've been r^(n-1) in the first equation and therefore r^n in the second.
a!
I the LORD search the heart, I try the reins, even to give every man according to his ways, and according to the fruit of his doings.
✝️Jeremiah 17:10 KJV✝️
Jeremiah is dead?
He definitely made an error here - in the top series, he is starting at exp-0 which means the terms span 0-thru-N and the # of terms equals N+1. In the second series, he starts at exp-1 which means the terms span 1-thru-N and the # of terms equals N for a difference of 1 between the two series. This is where he made his error - He is trying to cancel out N+1 terms on top with only N terms on bottom and you simply can't do that.
The correct formula is --> a(1 - r^N) / (1 - r)
If you still don't believe me, then test it as follows. Set all the parameters to simple numbers such as a=2, r=4, N=3 and then solve both formulas. Once you've done that, then solve the series manually by actually hand-writing the series for each N and compare that solution with the two previous results - You will then see his error. Let me illustrate:
PARAMETERS: a=2; r=4; N=3
KHAN'S FORMULA: a(1 - r^N+1) / (1 - r) --> 2(1 - 4^4) / 1 - 4 --> -510 / -2 = 255
CORRECT FORMULA: a(1 - r^N) / (1 - r) --> 2(1 - 4^3) / 1 - 4 --> -126 / -3 = 42
MANUAL CALC: a(r^0) + a(r^1) + a(r^2) = 2(1) + 2(4) + 2(16) --> 2 + 8 + 32 = 42 - same as the "correct" formula above.
Another reason some folks get confused is the parameter N and the actual exponents - These are NOT interchangeable. N is the number of terms - NOT the max exponent. So in my simplistic example, the highest term is not r^N - it is r^N-1 or r^2.
cyberzeus agree
I agree on this
ahhh, I get the difference, but I don't get why khan's wrong. he's not exactly cancelling it out, he's subtracting. I think the definition of n is just different.
khan's manual calc would look like this: a(r^0) + a(r^1) + a(r^2) + a(r^3)
= 2 + 2(4^1) + 2(4^2) + 2(4^3) = 170
and 2(1 - 4^4) / (1 - 4) =170
I don't know how you got 255
n is not a max exponent, but it's not a number of terms either, it's a last "index". If k=999 and n=1000, the series will have two terms, not a thousand. And when k=0 and n=n, the number of terms will be n+1 and the last term will be a^n.
And the second series is not k=1 n=n, so why would it have less terms? The number of terms is still n+1, he's just raising r to k+1 power.
While the canonic formula for geometric series is indeed a(1-r^n)/(1-r), it is only because the last index is assumed to be n-1, and not n. So Khan didn't made a mistake here, he just uses different number of terms. Though he should be denoting S with n+1 then, and not n.
The n he's using is the last exponent not the number of terms
I know it works, but why? Why does x(1-r) work?
+Kevin Wang Why does anything work in math? Honestly, the farther I delve into the concepts, the less I understand.
So for rSn, if you take the r out, it'll just be: r x (ar^0+ar^1+ar^2+ar^3...)
And since Sn =(ar^0+ar^1+ar^2+ar^3...)
Then Sn- rSn= (ar^0+ar^1+ar^2+ar^3...) -r x (ar^0+ar^1+ar^2+ar^3...)
And you can simplify that to
(ar^0+ar^1+ar^2+ar^3...)x(1-r)
So what he's saying is that
(ar^0+ar^1+ar^2+ar^3...)x(1-r)= a-ar^n+1
And that solves to
(ar^0+ar^1+ar^2+ar^3...)= (a-ar^n+1)/(1-r)
We need IMO Solution in videos
The International Mathematical Olympiad (IMO)
3 - 3 = 6
Well fuck
this is precalculus?
yas
This is calc 2, not pre-calc lol
Guess that depends, I had this on my university entrance exam lol
This is super confusing and you need to pick better colors and write bigger. I can hardly see this
use full screen and a higher quality