Chapter 10.1, Problem 50E

To determine

To find: The sum of all the squares.

Answer to Problem 50E

Total area of squares $=8m^2$

Explanation of Solution

Given information:

The area of outermost square = 4 m².

Calculation:

Let the area of outermost be $A_n{A}_n=4\text{sq}.$ Therefore, side of the square $a_n=2\text{m}$

Area of preceding square is $A_{n-1}$ . Consider the right angle triangle at corner obtained by outermost square and it's preceding square. Where ${\left(\frac{a_n}{2}\right)}^2+{\left(\frac{a_n}{2}\right)}^2=a_{n-1}^2{a}_{n-1}=\sqrt{2}$

Therefore, $A_{n-1}={(\sqrt{2})}^2=2$ . Similarly, $A_{n-2}=1$ $\text{m}{A}_{n-3}=\frac{1}{2}$ and so on....

Let the sum of all areas be $AA=A_n+A_{n-1}+A_{n-2}+A_{n-3}+\dots \dots \mathrm{...}$

  $A=4+2+1+\frac{1}{2}+\frac{1}{4}+\dots \dots .$ The sum of the infinite geometric series whose ratio is $<1$ .

  $S=\frac{a}{1-r}$ where, $a$ is initial term and $r$ is the common ratio. It is a geometric progression with initial term is 4 and common ratio is equal to $\frac{1}{2}.A=\frac{4}{1-\frac{1}{2}}=8$

Total area of squares $=8m^2$