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Data Str &amp; Algorithm HW Solutions 12

# Data Str &amp; Algorithm HW Solutions 12 - n holes...

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12 Chap. 2 Mathematical Preliminaries = 8 3 ( 5 3 ) n 2 < ( 5 3 ) 2 ( 5 3 ) n 2 = 5 3 n . Thus, the theorem is proved by mathematical induction. 2.24 Proof : (a) Base case. For n = 1 , 1 3 = 1 2 (1+1) 2 4 = 1 . Thus, the formula is correct for the base case. (b) Induction Hypothesis. n 1 i =0 i 3 = ( n 1) 2 n 2 4 . (c) Induction Step. n i =0 i 3 = ( n 1) 2 n 2 4 + n 3 = n 4 2 n 3 + n 2 4 + n 3 = n 4 + 2 n 3 + n 2 4 = n 2 ( n 2 + 2 n + 2) 4 = n 2 ( n + 1) 2 4 Thus, the theorem is proved by mathematical induction. 2.25 (a) Proof : By contradiction. Assume that the theorem is false. Then, each pigeonhole contains at most 1 pigeon. Since there are
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Unformatted text preview: n holes, there is room for only n pigeons. This contradicts the fact that a total of n + 1 pigeons are within the n holes. Thus, the theorem must be correct. ✷ (b) Proof : i. Base case. For one pigeon hole and two pigeons, there must be two pigeons in the hole. ii. Induction Hypothesis. For n pigeons in n − 1 holes, some hole must contain at least two pigeons....
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