Finite geometric series formula derivation? why r*S?

In summary, multiplying "r" to the second line of series and then cancelling those terms gives us a valid, sound, and logical answer. This is because of the process of distributing (1 - r) to the polynomial, which gives us 1*S - r*S = S - rS, leading to the desired result.
  • #1
Terrell
317
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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
 
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  • #2
Terrell said:
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.
 
  • #3
Samy_A said:
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.
Samy_A said:
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.
 
  • #4
Terrell said:
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.
 

FAQ: Finite geometric series formula derivation? why r*S?

1. What is a finite geometric series?

A finite geometric series is a sequence of numbers where each term is obtained by multiplying the previous term by a constant number, called the common ratio. It has a fixed number of terms and can be represented as a sum of terms in the form of a geometric progression.

2. What is the formula for finding the sum of a finite geometric series?

The formula for finding the sum of a finite geometric series is S = a * (1 - r^n) / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms in the series.

3. How is the formula for a finite geometric series derived?

The formula for a finite geometric series is derived using the concept of geometric progression, where each term is multiplied by the common ratio to get the next term. By summing up all the terms, a pattern is observed which leads to the formula S = a * (1 - r^n) / (1 - r).

4. Why is the common ratio (r) included in the formula for a finite geometric series?

The common ratio (r) is included in the formula because it represents the rate at which each term in the series is increasing or decreasing. It is a crucial factor in determining the sum of the series and cannot be ignored in the formula.

5. What is the significance of the 'r*S' term in the formula for a finite geometric series?

The 'r*S' term in the formula represents the sum of the series multiplied by the common ratio. It is included to show the relationship between the sum and the common ratio, and how it affects the overall value of the series. It also helps in understanding the concept of geometric progression better.

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