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Unformatted text preview: Math 131A  Section 2 Spring 2010 Homework 1 The following problems were graded for this assignment: 1.3 (2 points), 1.9 (3 points), 1.12 (3 points), 3.3 (2 points), 3.4 (2 points), and 3.6 (3 points), giving a total of 15 points. Below are some solutions as well as notes on common errors that were made (in the problems that were graded). 1.3. Let P n be the proposition that 1 3 + 2 3 + ··· + n 3 = (1 + 2 + ··· + n ) 2 . Since 1 3 = 1 2 , we have that P 1 is true. Suppose P n is true for some n ≥ 1. Then, using example 1 (the fact that 1 + ··· + n = n ( n + 1) / 2), we obtain: (1 + 2 + ··· + n + ( n + 1)) 2 = (1 + ··· + n ) 2 + 2( n + 1)(1 + ··· + n ) + ( n + 1) 2 = 1 3 + ··· + n 3 + 2( n + 1) n ( n + 1) 2 + ( n + 1) 2 = 1 3 + ··· + n 3 + n ( n + 1) 2 + ( n + 1) 2 = 1 3 + ··· + n 3 + ( n + 1) 3 , so P n +1 is true. By induction, we conclude that P n is true for all n ∈ N . Notes: If you are using some proposition called P n in your solution, make sure you explicitly write down what it says. There is no P n referenced in the statement of the problem, and while it might be obvious what it should be, that does not mean that it should be skipped over in your solution. Another common error was to take the equation (1 + 2 + ··· + n + ( n + 1)) 2 = 1 3 + 2 3 + ··· + n 3 + ( n + 1) 3 , and manipulate both sides of this equation to get something else that is true. When doing induction, try to work on one side of the equation and slowly transform it into the other side. You may have to write down some scratch work on the side in order to do it like this, but you should make a habit of it. Finally, almost everyone used the result 1 + 2 + ··· + n = n ( n + 1) / 2. When you have something like this in your solution, you should make note of it and where it came from (for example, “we know 1 + 2 + ··· + n = n (...
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This note was uploaded on 11/20/2010 for the course MATH 131A 131A taught by Professor Kim during the Spring '10 term at UCLA.
 Spring '10
 Kim

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