I need to solve this problem with two or more series or sequ

I need to solve this problem with two (or more) series or sequences that lead to the same solution

Solution

Let the side of Bigger Red square is of length a

Then length of next green square would be a/2

The ;length of next red square is a/4 and so on...

The Area of red squares would be given by :

a² + (a/4)² + (a/16)² + ...

= a² + a²/16 + a²/256+....

= a² [ 1+1/16+1/256+..]

= This is gemetric series with r = 1/16 and a = 1

The sum of infinite gemetric series is given by : a/(1-r) = 1/(1-1/16) = 16/15

Hence total area of red squares would be 16a²/15

Now consider the area of green squares is given by :

(a/2)²+(a/8)²+(a/32)²+...

= a²/4 + a²/64 + a²/1024+...

= a²/4(1+1/16+1/256)

This is same series as the previous series and hence the sum of this is 16/15

So the total area of green squares is given by 16a²/4.15 = 4a²/15

The ratio is then given by :

(16a²/15)/(4a²/15) = 4