# Problem with Area and Sequences/Series

• Sep 20th 2007, 12:47 PM
Norman Smith
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!

http://img204.imageshack.us/img204/5...problemmt3.jpg
• Sep 21st 2007, 12:22 AM
Spimon
Does the text book have an answer? Below is my solution. I hope it's right!

http://img227.imageshack.us/img227/2588/sequencegr8.jpg

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

I hope thats right.
• Sep 21st 2007, 09:24 AM
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
• Sep 21st 2007, 11:00 AM
CaptainBlack
Quote:

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
• Sep 21st 2007, 04:28 PM
ScottO
I aways seem to have a knack for finding a route that's a bit more complicated. (Doh)

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