Formula for a finite geometric series | Sequences, series and induction | Precalculus | Khan Academy

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[eveykaye]

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) 

[comicstwisted]

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

[AdnanHussain]

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.

[thegoonist]

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?

[codyelhard]

You make this easier to understand than the formula itself, thanks much!

[picsoumag]

why so we even go to school is it that simple? waw I'm amazed thank you so much

[obinator]

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.

[hasinanjum]

this video is missed in the series(Precalculas) course. Hopping that it will be

[melainamathgenius]

Again you are an awesome teacher. I kind of understand this because I am only in sixth grade.

[lorenzfuentes]

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.

[mshirazkhan]

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

[blengulilat]

The formula of the sum of geometric progression is a-ar^n / 1-r . Not the power of n+1

[josephjoe]

@2:43 why was r multiplied to Sn? Also, why was rSn subtracted to Sn?

[JeanAlesiagain]

This formula seems different to what I have seen in other videos: (a(1-r^n))/1-r
What's the difference?

[harryr]

I'm pretty sure that it should be a(1-r^n)/(1-r) and not a(1-r^(n+1))/(1-r)

[Joe-yoop]

Hey, can you tell me, why you could shift the rSn by one?

[seprage]

The question here is; why multiply r times Sn. Whats the purpose of doing it?

[huangweicheng]

why doesn't subtract the formula further to Sn=ar^n by cancelling (1-r)?

[dark]

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.

[mohanlalchoudhary]

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? :)