Adv Alegbra HW Solutions 4

Adv Alegbra HW - he had taken to the hospital was not an interesting number Ramanujan disagreed saying that it is the smallest positive integer

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4 1.9 Find a formula for 1 + n j = 1 j ! j , and use induction to prove that your formula is correct. Solution. A list of the sums for n = 1 , 2 , 3 , 4 , 5i s2 , 6 , 24 , 120 , 720. These are factorials; better, they are 2 ! , 3 ! , 4 ! , 5 ! , 6 ! . We have been led to the guess S ( n ) : 1 + n X j = 1 j ! j = ( n + 1 ) ! . We now use induction to prove that the guess is always true. The base step S ( 1 ) has already been checked; it is on the list. For the inductive step, we must prove S ( n + 1 ) : 1 + n + 1 X j = 1 j ! j = ( n + 2 ) ! . Rewrite the left side as h 1 + n X j = 1 j ! j i + ( n + 1 ) ! ( n + 1 ). By the inductive hypothesis, the bracketed term is ( n + 1 ) ! , and so the left side equals ( n + 1 ) !+ ( n + 1 ) ! ( n + 1 ) = ( n + 1 ) ![ 1 + ( n + 1 ) ] = ( n + 1 ) ! ( n + 2 ) = ( n + 2 ) ! . By induction, S ( n ) is true for all n 1. 1.10 ( M. Barr ) There is a famous anecdote describing a hospital visit of G. H.
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Unformatted text preview: he had taken to the hospital was not an interesting number. Ramanujan disagreed, saying that it is the smallest positive integer that can be written as the sum of two cubes in two different ways. (i) Prove that Ramanujan’s statement is true. Solution. First, 1729 is the sum of two cubes in two different ways: 1729 = 1 3 + 12 3 ; 1927 = 9 3 + 10 3 . Second, no smaller number n has this property. If n = a 3 + b 3 , then a , b ≤ 12. It is now a matter of checking all pairs a 3 + b 3 for such a and b ....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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