Visual Intuition for the Sum of a FINITE Geometric Series
I'm interested in intuitive visual explanations for the sum of a finite geometric series.
I know there are some pretty "intuitive" explanations out there (including some on this site), but I haven't seen any that provide a visual intuition.
If anyone here knows of any and would share them, I'd greatly appreciate it!
Thanks!
Finite geometric series. All the answers thus far have been for the infinite case.
Thanks!
$\endgroup$ 35 Answers
$\begingroup$Here's another proof (copied from brilliant.org) of the infinite series, but for arbitrary $r<1$. I wonder if it can be adapted for the finite case by thinking about a trapezoid like this instead of a triangle....
If anyone would survive being limited to natural numbers only, I suggest an illustration with the help of trees.
Each of the $n$ terms of a geometric series with a common ratio $r$ corresponds to a level in a (perfect) $r$-ary tree $T_n$. The sum $s$ is given by the number of nodes in $T_n$.
The main insight comes from comparing $T_n$ with $T_{n+1}$, i.e. comparing a series of $n$ and $n+1$ terms:
- Multiplying $T_n$ with $r$, we get $r$ replicas of $T_n$ that can be rooted in a dummy root, creating a tree $T_{n+1}$ with $sr + 1$ nodes.
- Now, we can cover the top part of $T_{n+1}$ by $T_{n}$ leaving only the leaves of $T_{n+1}$, i.e. $r^{n+1}$ nodes:$rs - s = r^{n+1} - 1$ (we take care of the annoying root by subtracting $1$)
- Another way to remove $s$ nodes from $T_{n+1}$ is to cover one of the replicas we created in step 1). This leaves us with $(r-1)$ replicas that should still consist of $r^{n+1} - 1$ nodes: $s(r-1) = r^{n+1} - 1$
- We're not interested in $(r-1)$ replicas, but only in one replica, so we divide by $r-1$ and hope that we were not drawing unary trees: $s=\frac{r^{n+1}-1}{r-1}$
In summary, by considering the next term of the series, you increase your sum by a known amount (step 2), which corresponds to multiplying the original by a known amount (step 3).
$\endgroup$ $\begingroup$See this images:
This are a graphic explaination of the sum of the geometric progression of ratio $\frac{1}{2}$.
$\endgroup$ 2 $\begingroup$Draw a segment one unit long.
Tick at the first third from the left.
On the right side, tick at the first third from the left.
On the right side, tick at the first third from the left.
On the right side, tick at the first third from the left.
…
When you are done, you have the infinite sum for $a=\frac13,r=\frac23$.
If you stop before infinity, the missing bit has the length $ar^n$.
$\endgroup$ $\begingroup$I think it is Matteo's method in disguise , since you want it for finite sum . (Very sorry for bad drawing) k is common difference (r).
Assume you want to calculate $$S=2+2^2+2^3+2^4$$.
Now imagine a rectangle of side 1 & 2 . Now take two more such rectangles and put them on adjacent positions they form a big rectangle of side 2 and 2 . now again take two big rectangle to repeat the process up to 4 times . (See diagram). Now all you need to is find the total area of these rectangles. Here two consecutive rectangles got one side common and area is twice of bigger.
Now lets go one step further in series and draw next rectangle which will have twice the area of last one . Now we want to check if this area is more then or less then S.
Now we try to fit our area in to this bigger rectangle.
Let divide it in to two parts one part of S fits in it we left with other part , again divide smaller one into two parts one part fits and we left with one smaller. Repeat this process our whole S fits in and we eventually left with area as much as smallest rectangle.
So $$\color{red}{S=2^5-2=32-2=30}$$
When you think about generalizing this method we need $\color{red}{k}$ more previous rectangles . Or we have to divide that one step further area in to k parts and we will use 1 part to cover up again , so we got k-1 parts to divide and we will eventually left with smallest rectangle. So the area is k-1 times.
$\endgroup$$$ S(k-1)=A_{n+1}-A_1$$
