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
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
- Fraction, Power of two, distinct linear factors