Intuition for the GP formula.

Question

Okay, so I wanted to ask, whether the formula for a geometric progression also has an intuition. There is an intuition for the AP formula: $$\frac n2(2a+(n-1)d)=\frac n2(a+(a+(n-1)d))$$ So what we are doing is, we are taking the average of the first term $a$ and $a+(n-1)d$ last term and multiplying it with the total number of terms. So we are doing kind of "scaling" to all the terms.
But I wonder if there is a good intuition for $\frac{a(1-r^n)}{1-r}$ formula, which could help me, if not remember, understand?
I have seen many people on this site, give beautiful solutions to these problems. Using diagrams etc. All this would surely be appreciated.

Answer 1

Sam and Tom decide to go on a diet for $n$ days, reducing their calorie intake to fraction $r$ of previous day's, starting with $a$, with the calendar day's calorie intake fixed as $ar^{(k-1)}$. Unfortunately, Tom starts a day late, so decides to just extend his diet by one day to complete his $n$ days.

Obviously the difference in total calorie intake $(S-T)$ between them is (Sam's first day's - Tom's last day's) = $(a - ar^n)$

But $(S-T)$ = S - Sr, so $S(1-r) = a(1-r^n)$, and the result follows.

The intuition here is to use the difference of extremes, in the A.P. it was to use the average of the extremes.

Answer 2

The geometric sum formula $S = \dfrac {a(r^n – 1)}{r – 1}$ can be re-written as $$S = a \cdot \dfrac {ar^n – a}{ar – a}$$ It can then be interpreted as

a times the ratio of ‘the difference between the (n+1)th term and 1st term’ to 'the difference between the 2nd term and 1st term’.

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Supplement

It is unfortunate to note that a geometric sum cannot be interpreted geometrically. That is, it is not quite possible to model the effect of translating an arithmetic sum to “n times the average of the first and the last term”.

After the above modification, the best I can do is to make the formula easier to be memorized. To this end, I have to makeup four new names:-

1. The strength between two (GS) terms = T(m) – T(n)

2. The basic strength $= T(2) – T(1) = ar – a$

3. The maximum strength $= T(n + 1) – T(1) = ar^ n – a$

4. R = The relative maximum strength = The maximum strength per unit of the basic strength= $ \dfrac { T(n + 1) – T(1)}{ T(2) – T(1)}$

Of course, it is neither easy to realize the geometrical effect of 'R' nor the geometric role of ‘a’ in the formula.

However, after the above transformation, we arrive at a formula that is much easier to memorize:- $$S = aR$$

Answer 3

First note that $$S_\infty=1+r+r^2+r^3+\cdots=\frac 1{1-r}$$ This can be shown by the binomial expansion of $(1-r)^{-1}$ or by synthetic division.

Now chop off terms from the $(n+1)$-th term to the rest of the series. $$\begin{align}&\overbrace{1+r+r^2+r^3+\cdots+r^{n-1}}^{\text{$n$ terms}}\\ =&\;\overbrace{1+r+r^2+r^3+\cdots+r^{n-1}+r^n+r^{n+1}+r^{n+2}+\cdots }^{S_\infty}\\ &\;\qquad\qquad\qquad\qquad\quad-r^n(\underbrace{1\;+r\;\;\;\;+r^2\;\;\;+\cdots}_{S_\infty})\\ =&\;S_\infty (1-r^n)\\\\ a+ar+ar^2+\cdots+ar^{n-1}=&\;a\left(1+r+r^2+r^3+\cdots+r^{n-1}\right) \\ =&\frac {a(1-r^n)}{1-r}=\frac {a(r^n-1)}{r-1}\end{align}$$

Answer 4

I like to think of the case where $ r=10$. Then the geometric series formula tells us, for example, that \begin{equation} 999 = 1000 - 1. \end{equation}

Answer 5

If $r$ is a positive integer, then a couple of ways to think of the formula (both covered in earlier answers in some way). The formula: $1+r+\ldots r^{n-1}=\frac{r^n-1}{r-1}$.

Intuition 1:

In base 10, we have 9 times $(1+10+\ldots 10^{n-1})$ is 9 times (11...1)=(99...9)=10^n-1. In base $r$, $(r-1) (1+r+\ldots r^{n-1})$ is just $\left((r-1) (r-1) \ldots (r-1) \right)$, which is just $r^n-1$.

Intuition 2:

Consider the case $r=2$. Now consider the number of games in a knockout tennis tournament played by $2^n$ players. The number of rounds is equal to the L.H.S., by counting the number of games for each round, and equal to the R.H.S. by observing that each round eliminates one player.

For $r$ being any positive integer, the same idea works: consider $r^n$ players and eliminate them in groups of size $r$. In each round, advance one person in each group.