Adv Alegbra HW Solutions 3

Adv Alegbra HW Solutions 3 - 1 5 ( n + 1 ) 5 + 1 2 ( n + 1...

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3 Since a is positive, Theorem 1.4(i) gives a n + 1 = aa n ab n ; since b is positive, Theorem 1.4(i) now gives ab n bb n = b n + 1 . 1.5 Prove that 1 2 + 2 2 +···+ n 2 = 1 6 n ( n + 1 )( 2 n + 1 ) = 1 3 n 3 + 1 2 n 2 + 1 6 n . Solution. The proof is by induction on n 1. When n = 1, the left side is 1 and the right side is 1 3 + 1 2 + 1 6 = 1. For the inductive step, [ 1 2 + 2 2 +···+ n 2 ]+ ( n + 1 ) 2 = 1 3 n 3 + 1 2 n 2 + 1 6 n + ( n + 1 ) 2 = 1 3 ( n + 1 ) 3 + 1 2 ( n + 1 ) 2 + 1 6 ( n + 1 ), after some elementary algebraic manipulation. 1.6 Prove that 1 3 + 2 3 +···+ n 3 = 1 4 n 4 + 1 2 n 3 + 1 4 n 2 . Solution. Base step : When n = 1, both sides equal 1. Inductive step : [ 1 3 + 2 3 +···+ n 3 ]+ ( n + 1 ) 3 = 1 4 n 4 + 1 2 n 3 + 1 4 n 2 + ( n + 1 ) 3 . Expanding gives 1 4 n 4 + 3 2 n 3 + 13 4 n 2 + 3 n + 1 , which is 1 4 ( n + 1 ) 4 + 1 2 ( n + 1 ) 3 + 1 4 ( n + 1 ) 2 . 1.7 Prove that 1 4 + 2 4 +···+ n 4 = 1 5 n 5 + 1 2 n 4 + 1 3 n 3 1 30 n . Solution. The proof is by induction on n 1. If n 1, then the left side is 1, while the right side is 1 5 + 1 2 + 1 3 1 30 = 1 as well. For the inductive step, h 1 4 + 2 4 +···+ n 4 i + ( n + 1 ) 4 = 1 5 n 5 + 1 2 n 4 + 1 3 n 3 1 30 n + ( n + 1 ) 4 . It is now straightforward to check that this last expression is equal to
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Unformatted text preview: 1 5 ( n + 1 ) 5 + 1 2 ( n + 1 ) 4 + 1 3 ( n + 1 ) 3 1 30 ( n + 1 ). 1.8 Find a formula for 1 + 3 + 5 ++ ( 2 n 1 ) , and use mathematical induction to prove that your formula is correct. Solution. We prove by induction on n 1 that the sum is n 2 . Base Step . When n = 1, we interpret the left side to mean 1. Of course, 1 2 = 1, and so the base step is true. Inductive Step . 1 + 3 + 5 + + ( 2 n 1 ) + ( 2 n + 1 ) = 1 + 3 + 5 + + ( 2 n 1 ) ] + ( 2 n + 1 ) = n 2 + 2 n + 1 = ( n + 1 ) 2 ....
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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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