If you’re wondering how you get the proof for finding Sn, that’s something that I think is very difficult to cover in a small internet comment. You really need some good pencil and paper to illustrate how it works.
But in a nutshell, you can list down all the terms:
Sn = a, ar, ar^2, … , ar^(n-2), ar^(n-1)
Then, you write down rSn (common ratio multiplied by Sn), so you get:
rSn = ar, ar^2, a^3, … , ar^(n-1), ar^n
Then you subtract rSn from Sn, and you can see that a lot of it cancels out [ ar, ar^2, … , ar^(n-1) ]
It’s a little hard to show in text form, but it makes more sense once you write it down. This leaves you with this:
Sn - rSn = a - ar^n
Sn(1 - r) = a(1 - r^n)
Then you divide by (1 - r) to get a simplified expression for Sn!
Sn = a(1 - r^n) / (1 - r)
And for the other equation (where absolute value of r > 1), you instead subtract Sn from rSn, then divide by (r - 1) instead of (1 - r). It’s the same logic though.
If you’re wondering how you get the proof for finding Sn, that’s something that I think is very difficult to cover in a small internet comment. You really need some good pencil and paper to illustrate how it works.
But in a nutshell, you can list down all the terms:
Sn = a, ar, ar^2, … , ar^(n-2), ar^(n-1)
Then, you write down rSn (common ratio multiplied by Sn), so you get:
rSn = ar, ar^2, a^3, … , ar^(n-1), ar^n
Then you subtract rSn from Sn, and you can see that a lot of it cancels out [ ar, ar^2, … , ar^(n-1) ]
It’s a little hard to show in text form, but it makes more sense once you write it down. This leaves you with this:
Sn - rSn = a - ar^n
Sn(1 - r) = a(1 - r^n)
Then you divide by (1 - r) to get a simplified expression for Sn!
Sn = a(1 - r^n) / (1 - r)
And for the other equation (where absolute value of r > 1), you instead subtract Sn from rSn, then divide by (r - 1) instead of (1 - r). It’s the same logic though.