Why does the geometric series formula intuitively work?

Question

I understand the proof for the geometric series formula, but I don't understand how the formula, $S_n = a_1\frac{(r^n -1)}{(r-1)}$

actually relates to the sum of all the terms.

What operations are taking place in the formula to give the sum.

Answer 1

I like to think of it for integers working over different bases.

For example, $9999 = 10000 -1$ is obvious and intuitive. Written another way, this reads: $$ 10^0 + 10^1 + \cdots + 10^3 = \frac{10^4 - 1}{10 -1} $$ Then just replace the $10$ with $x$.

Answer 2

Try subtracting these two,

$$\begin{align}S&=1+r+r^2+\dots+r^n\\-(\ rS&=\quad\ \ \ r+r^2+\dots+r^n+r^{n+1}\ )\\\hline(1-r)S&=1+0+0\phantom{^2}+\dots+0^\phantom{^n}-r^{n+1}\end{align}$$

So, we have

$$(1-r)S=1-r^{n+1}$$

$$S=\frac{1-r^{n+1}}{1-r}$$

So that we now have

$$1+r+r^2+\dots+r^n=\frac{1-r^{n+1}}{1-r}$$

Answer 3

We have a geometric sequence (or progression), in our case finite with $\;n\;$ terms:

$$a_1,\,a_1r,\,a_1r^2,\,\ldots,\,a_1r^{n-1}$$

Let's call the sum of the above sequence $\;S\;$ , so

$$\begin{align*}&S=\sum_{k=0}^{n-1}a_1r^k=a_1\left(1+r+\ldots+r^{n-1}\right)\implies\\{}\\&rS=a_1\left(r+r^2+\ldots+r^n\right)\implies\\{}\\&(1-r)S=a_1\left(1+r+\ldots+r^{n-1}-r-r^2-\ldots-r^{n-1}-r^n\right)=a_1(1-r^n)\implies\\{}\\&S=a_1\frac{1-r^n}{1-r}\end{align*}$$

Observe that in the above there exists hidden induction.

Answer 4

If ever in doubt, you can try induction. Notice that

$$a_1+a_1r=a_1\frac{1-r^2}{1-r}$$

So the formula is true for $n=1$.

Suppose that $a_1+a_1r+a_1r^2+\dots+a_1r^k=a_1\frac{1-r^{k+1}}{1-r}$ for some $k$. See then that

$$S=a_1+a_1r+a_1r^2+\dots+a_1r^k+a_1r^{k+1}\\=a_1\frac{1-r^{k+1}}{1-r}+a_1r^{k+1}\\=a_1\frac{1-r^{k+2}}{1-r}$$

This tells us that if the formula is true for $k=1:$

$$a_1+a_1r=a_1\frac{1-r^2}{1-r}$$

Then it is also true for $k+1:$

$$a_1+a_1r+a_1r^2=a_1\frac{1-r^3}{1-r}$$

And that for any $p=k+1$, it is also true for $p+1:$

$$a_1+a_1r+a_1r^2+a_1r^3=a_1\frac{1-r^4}{1-r}$$

etc.

Answer 5

The case $a_n=2^n$ is the most intuitive to me because:

$$2^n=2^n(2-1)=2^{n+1}-2^n$$

Then noticing the telescoping sum, the formula easily follows.

If we try to do the same kind off thing with $x$ instead of $2$ we get,

$$x^n=x^n\frac{x-1}{x-1}=\frac{x^{n+1}-x^n}{x-1}$$

$$=\frac{x^{n+1}}{x-1}-\frac{x^n}{x-1}$$

And again the result easily follows by telescoping.

Answer 6

Base n proof is the most intuitive way to reconstuct the formula.

Base 2 is easy

2^0 + 2^1 + 2^2 +... + 2^(n-1) = 1...11b = 2^n - 1

Base 5:

5^0 + 5^1 + 5^2 +... + 5^(n-1) = 1...11 base 5

     5^n - 1 = 4...44 base 5 
 (5^n - 1)/4 = 1...11 base 5

Base l| l is an integer:

 l^0 + l^1 + l^2 +... + l^(n-1) = 1...11 base l
                (l^n - 1)/(l-1) = 1...11 base l

Example:

    1 + 21 + 21^2 = 1 + 21 + 441 = 463
    (21^3 - 1)/20 = (9261 - 1)/20 = 463

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