Untitledv - C(k+1) . Therefore the area of C(k+1) must be...

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Suppose the area of the original square C(0) is equal to 1. To get C(k+1) , we scale C(k) by 1/3, which reduces the area by 1/9 = (1/3) 2 . But we make 8 copies of this scaled version to form
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Unformatted text preview: C(k+1) . Therefore the area of C(k+1) must be (8/9)th of the area of C(k) . This means that the area of C(n) is (8/9) n for all n ....
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This note was uploaded on 12/03/2009 for the course IAS 2045 taught by Professor Kent during the Spring '09 term at Arkansas State.

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