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115sols1 - HW 1 Solutions Math 115 Winter 2009 Prof Yitzhak...

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HW 1 Solutions Math 115, Winter 2009, Prof. Yitzhak Katznelson 1.1: Prove 1 2 + 2 2 + ... + n 2 = 1 6 n ( n + 1)(2 n + 1) for all natural numbers n . The proof is by induction. Call the n th proposition P n . The basis for induction P 1 is the statement that 1 2 = 1 6 1(1 + 1)(2 + 1), which is true. For the induction step, we assume that P n is true. We’d like to show that P n +1 is true, namely that: 1 2 + 2 2 + ... + n 2 + ( n + 1) 2 = 1 6 ( n + 1)(( n + 1) + 1)(2( n + 1) + 1) . By the induction step, the left-hand side is equal to 1 6 n ( n + 1)(2 n + 1) + ( n + 1) 2 ; we just replaced the first n terms by using P n . Then multiplying eerything out, we see that the left-hand side is: 1 6 (2 n 3 + 3 n 2 + n ) + n 2 + 2 n + 1 = 1 3 n 3 + 3 2 n 2 + 13 6 n + 1 . But the right-hand side is equal to: 1 6 ( n + 1)( n + 2)(2 n + 3) = 1 3 n 3 + 3 2 n 2 + 13 6 n + 1 , and so the left and right hand sides are equal. Thus P n +1 is true. By mathematical induction, this completes the proof. squaresolid 1.3: Prove 1 3 + 2 3 + ... + n 3 = (1 + 2 + ... + n ) 2 for all natural numbers n . The proof is again by induction; call the n th proposition P n . The basis for induction P 1 is the statement that 1 3 = 1 2 , which is obviously true. For the induction step, we assume that P n is true. We’d like to show that P n +1 is true, namely that: 1 3 + 2 3 + ... + n 3 + ( n + 1) 3 = (1 + 2 + ... + n + ( n + 1)) 2 . The right-hand side may be expanded using the usual formula ( a + b ) 2 = a 2 + 2 ab + b 2 , with a = 1 + 2 + ... + n and b = n + 1. We see that the right-hand side is equal to: (1 + 2 + ... + n ) 2 + 2(1 + 2 + ... + n )( n + 1) + ( n + 1) 2 . We use P n (applied to the first term) to write this as: (1 3 + 2 3 + ... + n 3 ) + 2(1 + 2 + ... + n )( n + 1) + ( n + 1) 2 . 1
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2 Then we use the result of Example 1 to write this as: (1 3 + 2 3 + ... + n 3 ) + 2( 1 2 n ( n + 1))( n + 1) + ( n + 1) 2 . Expanding out the last two terms and combining them, we see that they are equal to n 3 + 3 n 2 + 3 n + 1 = ( n + 1) 3 , so our right-hand side is equal to: (1 3 + 2 3 + ... + n 3 ) + ( n + 1) 3 , and so we’ve shown P n +1 is true. By mathematical induction, this completes the proof. squaresolid 1.4: Evaluating 1 + 3 + ... + (2 n - 1): for n = 1 we get 1, for n = 2 we get 4, for n = 3 we get 9, and for n = 4 we get 16. This leads us to guess that 1 + 3 + ... + (2 n
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