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Expert-verified Found in: Page 810 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # Fibonacci Sequence: Let${u}_{n}=\frac{{\left(1+\sqrt{5}\right)}^{n}-{\left(1-\sqrt{5}\right)}^{n}}{{2}^{n}\sqrt{5}}$define 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.

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

${u}_{1}=1&{u}_{2}=1$

(b) ${u}_{n+2}={u}_{n+1}+{u}_{n}$

(c) {un } is a Fibonacci series.

See the step by step solution

## Step 1. Write the given information.

The nth term of the sequence is:

${u}_{n}=\frac{{\left(1+\sqrt{5}\right)}^{n}-{\left(1-\sqrt{5}\right)}^{n}}{{2}^{n}\sqrt{5}}$

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

First term is:

${u}_{1}=\frac{{\left(1+\sqrt{5}\right)}^{1}-{\left(1-\sqrt{5}\right)}^{1}}{{2}^{1}\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{1}=\frac{1+\sqrt{5}-1+\sqrt{5}}{2\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{1}=\frac{2\sqrt{5}}{2\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{1}=1$

Second term is:

${u}_{1}=\frac{{\left(1+\sqrt{5}\right)}^{2}-{\left(1-\sqrt{5}\right)}^{2}}{{2}^{2}\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{2}=\frac{\left(1+2\sqrt{5}+5\right)-\left(1-2\sqrt{5}+5\right)}{4\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{2}=\frac{4\sqrt{5}}{4\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{2}=1$

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

Firstly, right hand side:

${u}_{n+2}=\frac{{\left(1+\sqrt{5}\right)}^{n+2}-{\left(1-\sqrt{5}\right)}^{n+2}}{{2}^{n+2}\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{n+2}=\frac{{\left(1+\sqrt{5}\right)}^{n}{\left(1+\sqrt{5}\right)}^{2}-{\left(1-\sqrt{5}\right)}^{n}{\left(1-\sqrt{5}\right)}^{2}}{{2}^{2}×{2}^{n}\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{n+2}=\frac{{\left(1+\sqrt{5}\right)}^{n}\left[1+5+2\sqrt{5}\right]-{\left(1-\sqrt{5}\right)}^{n}\left[1-2\sqrt{5}+5\right]}{4×{2}^{n}\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{n+2}=\frac{2×\left\{{\left(1+\sqrt{5}\right)}^{n}\left[3+\sqrt{5}\right]-{\left(1-\sqrt{5}\right)}^{n}\left[3-\sqrt{5}\right]\right\}}{4×{2}^{n}\sqrt{5}}\phantom{\rule{0ex}{0ex}}⇒{u}_{n+2}=\frac{\left\{{\left(1+\sqrt{5}\right)}^{n}\left[3+\sqrt{5}\right]-{\left(1-\sqrt{5}\right)}^{n}\left[3-\sqrt{5}\right]\right\}}{2×{2}^{n}\sqrt{5}}$

Now, the left hand side:

${u}_{n+1}+{u}_{n}=\left[\frac{{\left(1+\sqrt{5}\right)}^{n+1}-{\left(1-\sqrt{5}\right)}^{n+1}}{{2}^{n+1}\sqrt{5}}\right]+\left[\frac{{\left(1+\sqrt{5}\right)}^{n}-{\left(1-\sqrt{5}\right)}^{n}}{{2}^{n}\sqrt{5}}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\sqrt{5}\right){\left(1+\sqrt{5}\right)}^{n}-{\left(1-\sqrt{5}\right)\left(1-\sqrt{5}\right)}^{n}+2{\left(1+\sqrt{5}\right)}^{n}-2{\left(1-\sqrt{5}\right)}^{n}}{2×{2}^{n}\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\sqrt{5}\right){\left(1+\sqrt{5}\right)}^{n}+2{\left(1+\sqrt{5}\right)}^{n}-{\left(1-\sqrt{5}\right)\left(1-\sqrt{5}\right)}^{n}-2{\left(1-\sqrt{5}\right)}^{n}}{2×{2}^{n}\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\left[1+\sqrt{5}+2\right]{\left(1+\sqrt{5}\right)}^{n}-{\left[1-\sqrt{5}+2\right]\left(1-\sqrt{5}\right)}^{n}}{2×{2}^{n}\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\left[3+\sqrt{5}\right]{\left(1+\sqrt{5}\right)}^{n}-{\left[3-\sqrt{5}\right]\left(1-\sqrt{5}\right)}^{n}}{2×{2}^{n}\sqrt{5}}$

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, {un} is a Fibonacci sequence. ### Want to see more solutions like these? 