Finite geometric series formula derivation? why r*S?

[Terrell]

what is the rationale of multiplying "r" to the second line of series? why does cancelling those terms give us a VALID, sound, logical answer? please help. here's a video of the procedure

 

[Samy_A]

what is the rationale of multiplying "r" to the second line of series? why does cancelling those terms give us a VALID, sound, logical answer? please help. here's a video of the procedure

The "rationale" is that it works.

Why wouldn't that operation be valid?
What he basically does is:

$a=b+c+d+e$
$f=\ \ \ \ \ \ c+d+e+g$
then, subtracting the second equation from the first gives $a-f=b-g$

The other terms cancel each other.

 

[Terrell]

The "rationale" is that it works.

Why wouldn't that operation be valid?
What he basically does is:

$a=b+c+d+e$
$f=\ \ \ \ \ \ c+d+e+g$
then, subtracting the second equation from the first gives $a-f=b-g$

The other terms cancel each other.

The "rationale" is that it works.

Why wouldn't that operation be valid?
What he basically does is:

$a=b+c+d+e$
$f=\ \ \ \ \ \ c+d+e+g$
then, subtracting the second equation from the first gives $a-f=b-g$

The other terms cancel each other.

i get that it works, but the closest logical explanation i have for myself is 1 - (r^n) divided by 1 - r is the polynomial 1 + r + r^2 + ... + r^(n-1) + r^n. which is identical to the finite geometric series. Thus, distributing (1 - r) to the polynomial gives us 1*S -r*S = S - rS.

 

[Samy_A]

i get that it works, but the closest logical explanation i have for myself is 1 - (r^n) divided by 1 - r is the polynomial 1 + r + r^2 + ... + r^(n-1) + r^n. which is identical to the finite geometric series. Thus, distributing (1 - r) to the polynomial gives us 1*S -r*S = S - rS.

I see.
Yes, that is correct.