Arithmetic Sequences: ua-cam.com/video/mxEIFO8vXKY/v-deo.html Geometric Sequences: ua-cam.com/video/zJEWcysAc44/v-deo.html Final Exams and Video Playlists: www.video-tutor.net/
I get that the General Formula works, and how it is derived. but not sure why it is multiplied by r (or -r). Why did someone decide to then add the 2 formulas. The arthmetic series formula and how it is derived makes so much more sense.
@@wyattdelaney-lefebvre1561just started learning about this aswell and all the proofs I see jjst randomly throw in subtracting rSn from Sn. Did you ever find an answer to that?
This is so funny I was looking for the reason as well and every video I find they just do it without an explanation. And people just soak it up. I'm glad I'm not crazy
its a trick to increase the power of r of each term on the right side by 1, so now when u minus the two things the terms cancel out and ur left with only a few Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 -r*Sn = a^r+ar^2+ar^4 ... ar^N when u add these both series u willl get Sn - r*Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 - (a^r+ar^2+ar^4 ... ar^N) terms on the right cancel out and u are left with Sn(1-r) = a-ar^N Sn = a(1-r^N) / (1-r)
its a trick to increase the power of r of each term on the right side by 1, so now when u minus the two things the terms cancel out and ur left with only a few Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 -r*Sn = a^r+ar^2+ar^4 ... ar^N when u add these both series u willl get Sn - r*Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 - (a^r+ar^2+ar^4 ... ar^N) terms on the right cancel out and u are left with Sn(1-r) = a-ar^N Sn = a(1-r^N) / (1-r)
It is one of those algebraic tricks that are used in mathematical proofs to derive a formula for something that is not too obvious to see. Herein, the infinite series is very difficult to see because it is infinite. So this is our algebraic manipulation to find the sum. It is genius and not every one could think about it.
its a trick to increase the power of r of each term on the right side by 1, so now when u minus the two things the terms cancel out and ur left with only a few Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 -r*Sn = a^r+ar^2+ar^4 ... ar^N when u add these both series u willl get Sn - r*Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 - (a^r+ar^2+ar^4 ... ar^N) terms on the right cancel out and u are left with Sn(1-r) = a-ar^N Sn = a(1-r^N) / (1-r)
I was really confused as to why the series was multiplied by r. I found another video (and webpage) that explains this better, which I'll reply to this comment with in case the channel owner wants to delete it (I get it, I am techichally advertising another creator) so he doesnt delete my explanation too. !!(MY EXPLANATION IS IN THIS COMMENTS REPLIES)!!
Basically, by multiplying by r, you "shifted over the summation, and got one more new term: r^n." This is important in the next step, because subtracting the two equations (rSn from Sn, or Sn-rSn) cancels out all but two terms: the a1, which was 'lost' when you multiplied Sn by r, and the new term from rSn, r^n. That leaves you with "Sn - rSn = a1-r^n". From there, it's what's said in the video: Isolate Sn. --->(factor Sn out on the left side, getting Sn(1-r), then---> factor a1 out of the left side, getting a(1-r^n), then---> divide both sides by (1-r) to get the final result Sn=(a(1-r^n))/(r-1).
!! LOOK AT EDIT WHEN YOU READ THIS PART !! When looking up the formulas for the geometric sum formula, I was also really confused because there were a bunch of different versions of it, and nothing was very clear about why they were different, so here's a reminder: a((r^n-1)/(r-1)) = (a(r^n-1))/(r-1) because a is a/1, so you've just multiplied a out. Also: a((r^n-1)/(r^n-1)) = a((1-r^n)/(1-r)) because you're multiplying the fraction by -1/-1, meaning nothing is changing (the negatives cancel each other out, aka you're technically multiplying the fraction by 1). Just remember that both the numerator and the denominator have to be written the same, because -1/1 doesn't equal 1, and 1/-1 does not equal one. Edit: I just came across something on www.cuemath.com/geometric-sum-formula/ that says if |r|< 1, the sum formula for finite terms is a((1-r^n)/(1-r)), and if |r|>1, the sum formula for the finite terms is a((r^n-1)/(r^n-1)). If you can, I would ask your teacher/Tutor/anyone who is good with math about it.
Arithmetic Sequences: ua-cam.com/video/mxEIFO8vXKY/v-deo.html
Geometric Sequences: ua-cam.com/video/zJEWcysAc44/v-deo.html
Final Exams and Video Playlists: www.video-tutor.net/
Bro this guys is definitely a polymath, genius
I've got an exam 3 hours from now. Binge-watching your vids really helped me prep. Thank you sir, you're a legend!
My lecture actually left us with this question of how to derive the formula. I didn't even know where to start 😢. You are a life saver . Thank you
Amazing video, after watching so many of these, I've finally understood that sum of infinite series one!
Thank you so much sir!
This way of learning math is what I do love. Thanks for the amazing teaching!
Nice demonstration of the sum formula of a geometric series. The only demonstration I have known before implies polynomial divison.
1:00 minute of the video and I got the concept, thank you🙌🏼
Thanks a lot!
Thank you for posting It's very helpfull
11:50
Literally mind blowing
You are amazing, man
Thank you so much 😊
Keep the good job.
I was stucked again and here is my saviour❤❤❤❤
Thank you!
Thanks man 🙏✊
If u ever need help in math this guy will always pop up
Thank you so much ❤
super cool stuff ^.^ very interesting.
Congratulations 🎈🍾 for more than 3M subs
Thank you. 🌻
You are a legend !
Thank you
thanks
Petition to buy this man a new mic
+1
Genius!
Could you please tell us witch program do you use for totirial.
Good video
thank u
I like this
hello. Good day. Can I ask for permission from you to let use your video in my unit plan.
For educational purposes only.
How we can find the Polynomial
(Suppose it called P ) Where we find that function P is a composite of function P and P (PoP =P )🤯🤯??
thx
I get that the General Formula works, and how it is derived. but not sure why it is multiplied by r (or -r). Why did someone decide to then add the 2 formulas. The arthmetic series formula and how it is derived makes so much more sense.
I've been looking everywhere for an explanation for this
@@wyattdelaney-lefebvre1561just started learning about this aswell and all the proofs I see jjst randomly throw in subtracting rSn from Sn. Did you ever find an answer to that?
This is so funny I was looking for the reason as well and every video I find they just do it without an explanation. And people just soak it up. I'm glad I'm not crazy
Multiplication by r and subtracting the resulting equation is a way to find Sn by elimination the same way we do when solving simultaneous equations.
Why did you multiply sn with r??
its a trick to increase the power of r of each term on the right side by 1, so now when u minus the two things the terms cancel out and ur left with only a few
Sn = a + a^r +ar^2 + ar^3 ... ar^N-1
-r*Sn = a^r+ar^2+ar^4 ... ar^N
when u add these both series u willl get
Sn - r*Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 - (a^r+ar^2+ar^4 ... ar^N)
terms on the right cancel out and u are left with
Sn(1-r) = a-ar^N
Sn = a(1-r^N) / (1-r)
At 9:44 why do you add the two equations? What do those last three steps represent?
You can do whatever you want
One day I will be like you
14:05
what is |r|>1?
what drawing software do you use?
Probably Photoshop, but you can use anything like Krita
Anyone knows what app is he using?
Hello, may I ask why r is multiplied to the series...?
I’m guessing so that the n-1 exponent in the last term just becomes n.
its a trick to increase the power of r of each term on the right side by 1, so now when u minus the two things the terms cancel out and ur left with only a few
Sn = a + a^r +ar^2 + ar^3 ... ar^N-1
-r*Sn = a^r+ar^2+ar^4 ... ar^N
when u add these both series u willl get
Sn - r*Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 - (a^r+ar^2+ar^4 ... ar^N)
terms on the right cancel out and u are left with
Sn(1-r) = a-ar^N
Sn = a(1-r^N) / (1-r)
Sir you are so fucking cool I absolutely love your channel ❤❤❤
Its like AP ( ARITHMETIC PROGRESSION )
why is r^n as n approaches infinity equal to 0?
Because 0< r
*Why* do we minus rSn from Sn to get the formula. I don't have an intuitive understanding of it.
It is one of those algebraic tricks that are used in mathematical proofs to derive a formula for something that is not too obvious to see. Herein, the infinite series is very difficult to see because it is infinite. So this is our algebraic manipulation to find the sum. It is genius and not every one could think about it.
Did he remove some of his videos?
At 8:33 WHY do you multiply both sides by r?
Convenience
what' s the reason it's multiplied by r?
its a trick to increase the power of r of each term on the right side by 1, so now when u minus the two things the terms cancel out and ur left with only a few
Sn = a + a^r +ar^2 + ar^3 ... ar^N-1
-r*Sn = a^r+ar^2+ar^4 ... ar^N
when u add these both series u willl get
Sn - r*Sn = a + a^r +ar^2 + ar^3 ... ar^N-1 - (a^r+ar^2+ar^4 ... ar^N)
terms on the right cancel out and u are left with
Sn(1-r) = a-ar^N
Sn = a(1-r^N) / (1-r)
Bro study some algebra
@@Soul-cu8zn Your comment was really helpful.
@@malanhemal6574 good for you
😭Thankyouuuu
11:20 😮
Great, but for one little thing: you say "r sub n" when you mean "r raised to the power n," at the end of the video.
I was really confused as to why the series was multiplied by r. I found another video (and webpage) that explains this better, which I'll reply to this comment with in case the channel owner wants to delete it (I get it, I am techichally advertising another creator) so he doesnt delete my explanation too. !!(MY EXPLANATION IS IN THIS COMMENTS REPLIES)!!
Basically, by multiplying by r, you "shifted over the summation, and got one more new term: r^n." This is important in the next step, because subtracting the two equations (rSn from Sn, or Sn-rSn) cancels out all but two terms: the a1, which was 'lost' when you multiplied Sn by r, and the new term from rSn, r^n. That leaves you with "Sn - rSn = a1-r^n". From there, it's what's said in the video: Isolate Sn. --->(factor Sn out on the left side, getting Sn(1-r), then---> factor a1 out of the left side, getting a(1-r^n), then---> divide both sides by (1-r) to get the final result Sn=(a(1-r^n))/(r-1).
!! LOOK AT EDIT WHEN YOU READ THIS PART !! When looking up the formulas for the geometric sum formula, I was also really confused because there were a bunch of different versions of it, and nothing was very clear about why they were different, so here's a reminder: a((r^n-1)/(r-1)) = (a(r^n-1))/(r-1) because a is a/1, so you've just multiplied a out. Also: a((r^n-1)/(r^n-1)) = a((1-r^n)/(1-r)) because you're multiplying the fraction by -1/-1, meaning nothing is changing (the negatives cancel each other out, aka you're technically multiplying the fraction by 1). Just remember that both the numerator and the denominator have to be written the same, because -1/1 doesn't equal 1, and 1/-1 does not equal one.
Edit: I just came across something on www.cuemath.com/geometric-sum-formula/ that says if |r|< 1, the sum formula for finite terms is a((1-r^n)/(1-r)), and if |r|>1, the sum formula for the finite terms is a((r^n-1)/(r^n-1)). If you can, I would ask your teacher/Tutor/anyone who is good with math about it.
Hello
At 9:30 WHY do you multiply by -1?
10:08 this is where it truly begins
thx
Ok wow
Your place is in space X or NASA 👌
my brain is fucked!
comic sans.
later