Problem: A circle is inscribed in a square of side m, then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If \mathrm{S}_n} is the sum of the areas of the first n circles so inscribed, then, as n grows beyond all bounds, \mathrm{S}_n} approaches:
Answer Choices:
A. 2Οm2β
B. 83Οm2β
C. 3Οm2β
D. 4Οm2β
E. 8Οm2β
Solution:
The radius of the first circle is 21βm; the side of the second square is 2mβ2β; the radius of the second circle is 21β(2m2ββ)=22βmβ; and so forth. If S is the limiting value of Snβ, then S=Ο(2mβ)2+Ο(22βmβ)2+Ο(4mβ)2+β¦=Οm2(41β+81β+161β+β¦)=Οm2(21β).