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8. Use the non-recursive formula for the Fibonnaci numbers given in Theorem 3.3.7 to deduce that for all n, 2n-¹ divides (1) + ()+³¹()+..) +5 +5² 5 3

8. Use the non-recursive formula for the Fibonnaci numbers given in Theorem 3.3.7 to deduce that for all n, 2n-¹ divides (1) + ()+³¹()+..) +5 +5² 5 3

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 16E
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I need hel with understanding and solving question 8.

8. Use the non-recursive formula for the Fibonnaci numbers given in Theorem 3.3.7 to deduce that for all n, 2n-1 divides (₁) + (3) + ²() +-) +5 ..). 5
Transcribed Image Text:8. Use the non-recursive formula for the Fibonnaci numbers given in Theorem 3.3.7 to deduce that for all n, 2n-1 divides (₁) + (3) + ²() +-) +5 ..). 5
Theorem 3.3.7 A closed formula for the Fibonacci numbers f1, f2.... is: n /5 fn = √/13 ((² + ✓ ³ ) " - (₁² = ~³) "). 2 2 Proof: We use partial fractions to find a closed formula for fn, given the generating function g(x) from Theorem 3.3.5. = First, one may verify that that 1 - x - x² (1 ax) (1 - 3x), where a = (1 + √5)/2 and 3 = (1 - √5)/2. Then we write = = x g(x) = (1-ax)(1-3x) A B + 1-αx 1-3x A(18x) + B(1-ax) Xx (1-ax)(1-3x) g(x) X From this we deduce that A = 1/√5 and B = -A = -1/√5. So 1 1 1 √/35 (₁ G 1- Bx, Now let's use the series expansion for 1/(1-x) again, to write this as g(x) ((1+ ax + a²x² + ...) - (1 + Bx +3²x² + ...)) ((a-B)x+ (a²-3²) x² + ...). = = = Xx - - The result follows by considering the coefficient of xn.
Transcribed Image Text:Theorem 3.3.7 A closed formula for the Fibonacci numbers f1, f2.... is: n /5 fn = √/13 ((² + ✓ ³ ) " - (₁² = ~³) "). 2 2 Proof: We use partial fractions to find a closed formula for fn, given the generating function g(x) from Theorem 3.3.5. = First, one may verify that that 1 - x - x² (1 ax) (1 - 3x), where a = (1 + √5)/2 and 3 = (1 - √5)/2. Then we write = = x g(x) = (1-ax)(1-3x) A B + 1-αx 1-3x A(18x) + B(1-ax) Xx (1-ax)(1-3x) g(x) X From this we deduce that A = 1/√5 and B = -A = -1/√5. So 1 1 1 √/35 (₁ G 1- Bx, Now let's use the series expansion for 1/(1-x) again, to write this as g(x) ((1+ ax + a²x² + ...) - (1 + Bx +3²x² + ...)) ((a-B)x+ (a²-3²) x² + ...). = = = Xx - - The result follows by considering the coefficient of xn.
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