Geometric proof in Concrete Mathematics On page 32 of Concrete Mathematics by Graham, Knuth and Patashnik,  they demonstrate that the sum of a geometric progression is
$$
\sum_{k=0}^n a x^k = \frac{a-a x^{n+1}}{1-x}.
$$
In the margin next to this, there's a figure with the note "If it's geometric, there should be a geometric proof":

I don't understand what this picture is trying to show. Can anyone explain it?
 A: Consider the top edge of the diagram, and denote the rectangles that share an edge with this edge $R_1, R_2, \ldots$ in decreasing size. Similarly, let $S_1, S_2, \ldots$ denote the rectanges that share a right edge with the right edge of the diagram. Let $c$ and $d$ be the height and width of the diagram. Let $a$ be the area of $R_1\cup S_1$. Finally, let $\sqrt{x}$ be the ratio of the heights of $R_2$ and $R_1$, as well as the ratio of the widths of $S_1$ and $R_1$. Note that $cd - cdx = a$, so $cd = a/(1-x)$.
Continue the diagram to the $n+1$st iteration so we have $R_1, \ldots, R_{n+1}$ and $S_1, \ldots, S_{n+1}$. One the one hand, the area of $R_k\cup S_k$ is $ax^k$, so the total area is $\sum_{k=0}^nax^k$. Since the area of the left out rectangle is $cdx^{n+1}$, the area is also $cd - cdx^{n+1} = (a - ax^{n+1}) / (1-x)$
A: Let $n=3$, and $a>0$. Factor $a$ arbitrarily into positive base and height $bh=a$. In the picture we are given four hatched rectangles, which I call #1,#2,#3,#4 in decreasing size. But two additional rectangles are implied by the figure, which I call #5 and #6. Rectangle #5 is the small missing piece that is needed to finish the full rectangle #6. Rectangle #6 is shown in your picture to have $b<h$.
Now let $0<x<1/2.$ For convenience let $y=1-x$, so that $1/2<y<1$.
The biggest rectangle #6 has base $b$ and height $h$.
Rectangle #1 has the same height $h$ but smaller base $yb$.
Rectangle #2 has base $xb$ and height $yh.$
Rectangle #3 has base $xyb$ and height $xh.$
Rectangle #4 has base $x^2b$ and height $xyh.$
Rectangle #5 has base $x^2b$ and height $x^2h$. (I have decided whether to apply a factor of $x$ versus $y$ in each step by noticing (in your picture) if the new length is less than half of the old versus greater than half.)
The area $a=bh$ of rectangle #6 is the sum of the five areas of #1,#2,#3,#4,#5:
$$a = bh = yb(h)\,\,+\,\,xb(yh)\,\,+\,\,xyb(xh)\,\,+\,\,x^2b(xyh)\,\,+\,\,x^2b(x^2h)$$
$$ \,=\, bh\left(y\,\,+\,\,x(y)\,\,+\,\,xy(x)\,\,+\,\,x^2(xy)\,\,+\,\,x^2(x^2)\right)$$
$$=bhy\left(1\,\,+\,\,x\,\,+\,\,x^2\,\,+\,\,x^3\,\,+\,\,\frac{x^4}{y}\right).$$
This is
$$a=a(1-x)\left(1\,\,+\,\,x\,\,+\,\,x^2\,\,+\,\,x^3\,\,+\,\,\frac{x^4}{1-x}\right),$$
$$\frac{a}{1-x}=a(1\,\,+\,\,x\,\,+\,\,x^2\,\,+\,\,x^3)+a\left(\frac{x^4}{1-x}\right).$$
