A square has a side length of 2. The midpoints of each of the four sides are joined to form another square within the first square. The midpoints of the second square are then joined to form a third square within the second, and this process continues indefinitely. Find an expression for the perimeter of all of the squares combined.
Let $S_n$ be the perimeter of the $n$th square. It is easy to see that $S_1 = 8.$ We will prove for all $n \geq 1,$ $S_n = 2^{n+2}.$
[asy] size(8cm); D((0,0)--(1,0),linewidth(3)); D((1,0)--(2,0),linewidth(3)); D((2,0)--(3,0),linewidth(3)); D((3,0)--(4,0),linewidth(3)); D((0,1)--(0,0),linewidth(3)); D((0,0)--(0,-1),linewidth(3)); D((0,-1)--(0,-2),linewidth(3)); D((0,-2)--(0,-3),linewidth(3)); D((0,0)--(1/2,1/2)--(1,0),linewidth(1)); D((1/2,1/2)--(1/2,-1/2)--(-1/2,-1/2)--(-1/2,1/2)--(1/2,1/2)); D((1,1)--(1/2,1/2)--(1,-1/2)--(3/2,-1/2)--(3/2,1/2)--(1/2,1/2)); D((2,1)--(3/2,1/2)--(2,-1/2)--(5/2,-1/2)--(5/2,1/2)--(3/2,1/2)); D((3,1)--(5/2,1/2)--(3,-1/2)--(7/2,-1/2)--(7/2,1/2)--(5/2,1/2)); [/asy]
Clearly, the perimeter of the $n$th square is the perimeter of the $(n-1)$th square adding on $4(2^{-n+2})$ more. Then
\begin{align*} S_1 &= 8, \\ S_2 &= 8 + 4\left(\frac{1}{4}\right) = 10, \\ S_3 &= 10 + 4\left(\frac{1}{8}\right) = 10 + 4(2^{-3}) = \frac{42}{4} + \frac{4}{4} = \frac{46}{4}, \\ &\ \vdots \\ S_n &= \frac{2^{n+2} \cdot (2^{n+2} - 2)}{(2^2 - 1)} = \boxed{2^{n+2}}. \end{align*}
A square has a side length of 2. The midpoints of each of the four sides are joined to form another square within the first square. The midpoints of the second square are then joined to form a third square within the second, and this process continues indefinitely. Find an expression for the perimeter of all of the squares combined.
A)
B)
C)
D)
Let's first look at the first few squares formed by joining the midpoints of the sides of the original square.
[asy]
unitsize(10mm);
pair A = (0,0);
pair B = (2,0);
pair C = (2,2);
pair D = (0,2);
pair E = midpoint(A--B);
pair F = midpoint(B--C);
pair G = midpoint(C--D);
pair H = midpoint(D--A);
pair I = midpoint(E--F);
pair J = midpoint(F--G);
pair K = midpoint(G--H);
pair L = midpoint(H--I);
draw(A--B--C--D--cycle, linewidth(0.8));
draw(E--F--G--H--cycle, linewidth(0.8));
draw(I--J--K--L--cycle, linewidth(0.8), dashed);
label("$S_1$", (1, -0.3));
label("$S_2$", (1, 1.7));
label("$S_3$", (1, 3.7));
[/asy]
We can observe that the perimeter of each square is twice the perimeter of the previous square. Since the perimeter of the original square is $2(2+2+2+2) = 16$, the perimeter of the second square is $2(16) = 32$, the perimeter of the third square is $2(32) = 64$, and so on.
In general, let $P_n$ be the perimeter of the $n$th square. We have that $P_1 = 16$, and for each subsequent square, $P_n = 2P_{n-1}$. Since the perimeter of the $n$th square is twice the perimeter of the $(n-1)$th square, the perimeter of the $n$th square is $2^n$ times the perimeter of the first square.
Therefore, the perimeter of all the squares combined is the sum of the perimeters of all the squares, which is
\[P_1 + P_2 + P_3 + \ldots = 16 + 32 + 64 + \ldots = \sum_{n=0}^{\infty} 2^n = \frac{1}{1-2} = \boxed{-1}.\]We can also find the same answer more rigorously by using the formula for the sum of an infinite geometric series.