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Unformatted text preview: Section 20 2. Show that 1 + 2 + 3 + &&& + n = n ( n + 1) 2 for all positive integers n . The basis step is to verify that the equation holds for n = 1 . In this case the lefthand side is 1 and the righthand side is 1(1 + 1) = 2 = 1 . The induction step is to show that if the equation holds for n = k , then it holds for n = k + 1 . So we may assume (the induction hypothesis) that 1 + 2 + 3 + &&& + k = k ( k + 1) 2 Add k + 1 to both sides to get 1 + 2 + 3 + &&& + k + ( k + 1) = k ( k + 1) 2 + k + 1 The left hand side of this equation is what we get when we substitute k + 1 for n in the lefthand side of the given equation. The right hand side of this equation is equal to k ( k + 1) 2 + 2( k + 1) 2 = ( k + 2)( k + 1) 2 = ( k + 1)( k + 1 + 1) 2 which is what we get when we substitute k + 1 for n in the righthand side of the given equation. So we have shown that the given equation is true when n = k + 1 . 3. Show that n < 2 n for all n 2 N . As the smallest number in N is , the basis step is to verify that the inequality holds for n = 0 . Substituting for n gives < 2 = 1 , which is true. In the induction step , we assume that k < 2 k and prove that k + 1 < 2 k +1 . If we add 1 to both sides of k < 2 k we get k + 1 < 2 k + 1 : Note that 1 & 2 k for any k , so 2 k + 1 & 2 k + 2 k = 2 k +1 . Putting these two inequalities together gives k + 1 < 2 k +1 which is what we had to prove. 4. Show that n ! & n n for all positive integers n . The basis step is to verify that the inequality holds for n = 1 . Substi tuting 1 for n gives 1! & 1 1 , which is true because both sides are equal to 1 ....
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 Spring '08
 STAFF

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