How to make sense of the formula for the Sum of Geometric Series with finite terms?

Question

this is my first ever question here. My apologies if my formatting is a bit off, still learning. I am now studying the topic: geometric series with finite terms. I am having a very hard time relating the formula (that solves for the sum of the gemetric series) to the general equation for the geometric sequence.

So if this is the general equation for the geometric series, equation (A) below:

$(A): S_n = t_1 + (t_1 * r) + (t_1 * r^2) + (t_1 * r^3) + … + (t_1 * r^{n-2}) + (t_1 * r^{n-1})$

Where:

$t_1$ = our first term

$r$ = our multiplier (ratio between sequential terms)

$n$ = total number of terms

We modify equation (A) by multiplying both sides by $r$ to arrive at equation (B) below:

$(B): S_n * r = (t_1 * r) + (t_1 * r) * r + (t_1 * r^2) * r + (t_1 * r^3) * r + … + (t_1 * r^{n-2} * r) + (t_1 * r^{n-1} * r)$

And finally we arrive at equation (C) below:

$(C): S_n * r = t_1 * r + (t_1 * r^2) + (t_1 * r^3) + … + (t_1 * r^{n-2} + (t_1 * r^{n-1}) + (t_1 * r^n)$

I do understand how by multiplying both sides by $r$ we arrive at equation (C). I also understand how by subtracting equations (C) from (A) we arrive at equation (D) below:

$(D): S_n - S_n * r = t_1 - (t_1 * r^n)$, which we factor to arrive at $S_n * (1 - r) = t_1 * (1 - (r^n))$, which then becomes our final solution:

$$S_n = t_1 \frac{1 - r^n}{1 - r}$$

or its "mirror image" $S_n = t_1 \frac{r^n - 1}{r - 1}$

Here is my main question: I am having a difficult time INTUITIVELY connecting the solution

$$S_n = t_1\frac{r^n - 1}{r - 1}$$

to the original equation

$$S_n = t_1 + (t_1 * r) + (t_1 * r^2) + (t_1 * r^3) + … + (t_1 * r^{n-2}) + (t_1 * r^{n-1})$$

Can someone help me understand this intuitively? For example, I intuitively understand how Pi*(r^2)makes sense in solving for the area of a circle (by relating it to the area of a rectangle).

Here is my secondary question: How can we take the original equation (A), and subtract from it another equation we conveniently created?

I do understand that equation (C) is a legal derivative of equation (A), since we multiplied both sides by "r". But I am still struggling with the notion of subtracting a second equation (we made up) from the first equation in order to solve the first equation.

I am familiar with the method of Elimination to solve multiple equations, but in the case of Elimination Methodology the equation we eliminate is part of the given problem. In this case we didn't start with two equations, we started with a single equation and then made up the second equation.

What is this methodology called? I would like to look it up and study it further.

Also, what prevents us from solving the original equation (A) without resorting to a mathematical sleight of hand? Can't we solve it directly on its own merits?

Thanks for much in advance everyone. I have already learned so much from all of you in your various other discussions. I am eternally grateful to you all for sharing.

Answer 1

One way to get intuition is perhaps by noticing that we have the factorisation $$1-r^n = (1-r)(1+r+r^2+\cdots+r^{n-1}),$$ which is true because you can expand the right hand side yourself and see the magical cancellation happen. Once you get this, dividing both sides by $1-r$ gives exactly the geometric series summation formula.

For your secondary question, there's of course nothing wrong with manipulating a given statement to get another statement, then using both of these together. Logically, they're all consequences of the first thing you started from. I doubt there is a name for this "methodology" however, as it is just doing some clever algebraic manipulation.

As for the final question, I'm not sure there's going to be a satisfactory answer except to say that sometimes, when the answer to a question is not obvious, one needs to be a bit clever and see a nice idea in order to solve the problem. In this case, the nice idea is essentially the cancellation above, which is equivalent to the derivation you wrote in the question.