# Thread: Problem with Area and Sequences/Series

1. ## Problem with Area and Sequences/Series

Hi everyone! I was having this problem in textbook homework regarding sequences and series, but having to do with area which threw me off. There is a picture necessary so I have a picture of the question and picture to go with it. Thanks for all the help!

2. Does the text book have an answer? Below is my solution. I hope it's right!

This would mean after 9 repetitions, we would have 127.75 sq.in.

I hope thats right.

3. I think you need to figure a $\displaystyle \sqrt{2}$ into your series.

The sides of the triangles of the successive iterations are $\displaystyle \frac{1}{2}$ the hypotenuse of the earlier triangles forming the sides of the inner squares.

-Scott

4. Originally Posted by ScottO
I think you need to figure a $\displaystyle \sqrt{2}$ into your series.

The sides of the triangles of the successive iterations are $\displaystyle \frac{1}{2}$ the hypotenuse of the earlier triangles forming the sides of the inner squares.

-Scott
Succesive squares have side 1/sqrt(2) that of the preceeding square, so the
areas of corresponding similar sub areas decrease by a factor of 2 at each step.

RonL

5. I aways seem to have a knack for finding a route that's a bit more complicated.

My thinking was the sides of the squares were $\displaystyle \frac{s}{\sqrt{2}}$, so their areas were 1/2, and the areas of pairs of shaded triangles were 1/4.

So I put down this series...

$\displaystyle \frac{1}{4}\sum(\frac{s}{(\sqrt{2})^{n-1}})^2$

Which I see now, can be simplified to $\displaystyle \sum\frac{s^2}{2^{n+1}}$, the same thing Spimon shows above.

-Scott