# Adv Alegbra HW Solutions 10 - + 1 1 − β − 1 i´ = 1...

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10 If n 2, then F n = F n 1 + F n 2 < 2 n 1 + 2 n 2 (by inductive hypothesis) < 2 n 1 + 2 n 1 = 2 · 2 n 1 = 2 n . By induction, F n < 2 n for all n 0. Notice that the second form is the appropriate induction here, for we are using two predecessors, S ( n 2 ) and S ( n 1 ) , to prove S ( n ) . 1.19 If F n denotes the n th term of the Fibonacci sequence, prove that m X n = 1 F n = F m + 2 1 . Solution. By Theorem 1.15, we have F n = 1 5 n β n ) for all n . Hence, m X n = 1 F n = m X n = 1 1 5 n β n ) = 1 5 m X n = 1 n β n ) = 1 5 ³h 1 α m + 1 1 α 1 i h 1 β m + 1 1 β 1 . Now α(α 1 ) = 1, so that 1 /( 1 α) =− α ; similarly, 1 /( 1 β) =− β . Therefore, m X n = 1 F n = 1 5 ³h 1 α m + 1 1 α 1 i h 1 β m
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Unformatted text preview: + 1 1 − β − 1 i´ = 1 √ 5 ³ ( − α) [ ( 1 − α m + 1 ) − 1 ] − [ ( − β)( 1 − β m + 1 ) − 1 ] ´ = 1 √ 5 [− (α − β) + (α m + 2 − β m + 2 ) ] . This is the desired formula, for 1 √ 5 (α − β) = 1 and 1 √ 5 (α m + 2 − β m + 2 ) = F m + 2 . 1.20 Prove that 4 n + 1 + 5 2 n − 1 is divisible by 21 for all n ≥ 1. Solution. We use the second form of induction....
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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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