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Unformatted text preview: 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. Wed 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 lefthand 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 lefthand 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 righthand 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. Wed 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 righthand 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 righthand 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 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 ....
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This note was uploaded on 11/03/2010 for the course MATH 26233820 taught by Professor Gieseker,d. during the Fall '10 term at UCLA.
 Fall '10
 GIESEKER,D.
 Math, Algebra

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