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Unformatted text preview: Homework 2 Solutions Math 1a, 2007 Question 1a. Apostol page 45, problem 11: Determine all positive integers n for which 2 n < n ! Solution 1a. First we check that 2 n < n ! is false for n = 1 , 2 , 3, but that it is true for n = 4. 2 1 = 2 > 1 = 1! 2 2 = 4 > 2 = 2! 2 3 = 8 > 6 = 3! 2 4 = 16 < 24 = 4! Next, we prove by induction that 2 n < n ! holds for n ≥ 4. Base Case: shown above, for n = 4. Inductive Step: Assume 2 n < n ! holds for n = k , where k is a positive integer. Show it holds for n = k + 1. 2 k +1 = 2 · 2 k < 2 · k ! ≤ ( k + 1) · k ! (since k is a positive integer implies 2 ≤ k + 1) = ( k + 1)! So 2 n < n ! holds for all positive integers n with n ≥ 4 and it does not hold for n = 1 , 2 , 3. Question 1b. Apostol page 45, problem 12: a) Use the binomial theorem to prove that for n a positive integer we have 1 + 1 n n = 1 + n X k =1 1 k ! k 1 Y r =0 1 r n ! . Solution. 1 k ! k 1 Y r =0 n r n = 1 k ! n ( n 1) ... ( n ( k 1)) n k 1 = n !...
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This note was uploaded on 02/22/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.
 Fall '08
 Borodin,A
 Calculus, Linear Algebra, Algebra, Integers

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