Unformatted text preview: 58 1 Solutions . The last fraction has Thus a denominator with distinct linear factors so we get 31. Use Theorem 1 to compute the chain: where
and Continuing gives where
and
Thus . Now continue using Theorem 1 or replace by in the numerator of the last fraction to get
32. Use Theorem 1 to compute the where
and chain: ...
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 Fall '08
 BELL,D
 Fraction, Power of two, distinct linear factors

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