Book edition 6th

Author(s) Sullivan

Pages 1200 pages

ISBN 9780321795465

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Short Answer

Fibonacci Sequence: Let
$u_{n} = \frac{{(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5})}^{n}}{2^{n} \sqrt{5}}$
define the nth term of a sequence.
(a) Show that u₁= 1 and u₂ = 1.
(b) Show that u_n+2 = u_n+1 + u_n.
(c) Draw the conclusion that {u_n} is the Fibonacci sequence.

(a) The first (u₁) and second (u₂) term of the sequence are:

$u_{1} = 1 & u_{2} = 1$

(b) $u_{n + 2} = u_{n + 1} + u_{n}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Write the given information.

The n^thterm of the sequence is:

$u_{n} = \frac{{(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5})}^{n}}{2^{n} \sqrt{5}}$

Step 2. Use the nth sequence formula to calculate first and second term.

First term is:

$u_{1} = \frac{{(1 + \sqrt{5})}^{1} - {(1 - \sqrt{5})}^{1}}{2^{1} \sqrt{5}} \Rightarrow u_{1} = \frac{1 + \sqrt{5} - 1 + \sqrt{5}}{2 \sqrt{5}} \Rightarrow u_{1} = \frac{2 \sqrt{5}}{2 \sqrt{5}} \Rightarrow u_{1} = 1$

Second term is:

$u_{1} = \frac{{(1 + \sqrt{5})}^{2} - {(1 - \sqrt{5})}^{2}}{2^{2} \sqrt{5}} \Rightarrow u_{2} = \frac{(1 + 2 \sqrt{5} + 5) - (1 - 2 \sqrt{5} + 5)}{4 \sqrt{5}} \Rightarrow u_{2} = \frac{4 \sqrt{5}}{4 \sqrt{5}} \Rightarrow u_{2} = 1$

Step 3. To prove this expression, compute both sides of the expression to be equivalent.

Firstly, right hand side:

$u_{n + 2} = \frac{{(1 + \sqrt{5})}^{n + 2} - {(1 - \sqrt{5})}^{n + 2}}{2^{n + 2} \sqrt{5}} \Rightarrow u_{n + 2} = \frac{{(1 + \sqrt{5})}^{n} {(1 + \sqrt{5})}^{2} - {(1 - \sqrt{5})}^{n} {(1 - \sqrt{5})}^{2}}{2^{2} \times 2^{n} \sqrt{5}} \Rightarrow u_{n + 2} = \frac{{(1 + \sqrt{5})}^{n} [1 + 5 + 2 \sqrt{5}] - {(1 - \sqrt{5})}^{n} [1 - 2 \sqrt{5} + 5]}{4 \times 2^{n} \sqrt{5}} \Rightarrow u_{n + 2} = \frac{2 \times {{(1 + \sqrt{5})}^{n} [3 + \sqrt{5}] - {(1 - \sqrt{5})}^{n} [3 - \sqrt{5}]}}{4 \times 2^{n} \sqrt{5}} \Rightarrow u_{n + 2} = \frac{{{(1 + \sqrt{5})}^{n} [3 + \sqrt{5}] - {(1 - \sqrt{5})}^{n} [3 - \sqrt{5}]}}{2 \times 2^{n} \sqrt{5}}$

Now, the left hand side:

$u_{n + 1} + u_{n} = [\frac{{(1 + \sqrt{5})}^{n + 1} - {(1 - \sqrt{5})}^{n + 1}}{2^{n + 1} \sqrt{5}}] + [\frac{{(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5})}^{n}}{2^{n} \sqrt{5}}] = \frac{(1 + \sqrt{5}) {(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5}) (1 - \sqrt{5})}^{n} + 2 {(1 + \sqrt{5})}^{n} - 2 {(1 - \sqrt{5})}^{n}}{2 \times 2^{n} \sqrt{5}} = \frac{(1 + \sqrt{5}) {(1 + \sqrt{5})}^{n} + 2 {(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5}) (1 - \sqrt{5})}^{n} - 2 {(1 - \sqrt{5})}^{n}}{2 \times 2^{n} \sqrt{5}} = \frac{[1 + \sqrt{5} + 2] {(1 + \sqrt{5})}^{n} - {[1 - \sqrt{5} + 2] (1 - \sqrt{5})}^{n}}{2 \times 2^{n} \sqrt{5}} = \frac{[3 + \sqrt{5}] {(1 + \sqrt{5})}^{n} - {[3 - \sqrt{5}] (1 - \sqrt{5})}^{n}}{2 \times 2^{n} \sqrt{5}}$

Since both the sides have equal results, the expression stands valid.

Step 4. Use the expression from Step 3.

As proved in Step 3, the expression is a form of Fibonacci sequence. Thus, {u_n} is a Fibonacci sequence.

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Short Answer

Fibonacci Sequence: Let
$u_{n} = \frac{{(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5})}^{n}}{2^{n} \sqrt{5}}$
define the nth term of a sequence.
(a) Show that u₁= 1 and u₂ = 1.
(b) Show that u_n+2 = u_n+1 + u_n.
(c) Draw the conclusion that {u_n} is the Fibonacci sequence.

Step by Step Solution

Step 1. Write the given information.

Step 2. Use the nth sequence formula to calculate first and second term.

Step 3. To prove this expression, compute both sides of the expression to be equivalent.

Step 4. Use the expression from Step 3.

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Precalculus Enhanced with Graphing Utilities

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Short Answer

Fibonacci Sequence: Letun=(1+5)n-(1-5)n2n5define the nth term of a sequence.(a) Show that u1= 1 and u2 = 1.(b) Show that un+2 = un+1 + un.(c) Draw the conclusion that {un} is the Fibonacci sequence.

Step by Step Solution

Step 1. Write the given information.

Step 2. Use the nth sequence formula to calculate first and second term.

Step 3. To prove this expression, compute both sides of the expression to be equivalent.

Step 4. Use the expression from Step 3.

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Fibonacci Sequence: Let
$u_{n} = \frac{{(1 + \sqrt{5})}^{n} - {(1 - \sqrt{5})}^{n}}{2^{n} \sqrt{5}}$
define the nth term of a sequence.
(a) Show that u₁= 1 and u₂ = 1.
(b) Show that u_n+2 = u_n+1 + u_n.
(c) Draw the conclusion that {u_n} is the Fibonacci sequence.