Chapter 12.1, Problem 88AYU

(a)

To determine

The first $11$ terms of the Fibonacci sequence $u_{n+2}=u_{n+1}+u_n$ where the $n^{th}$ term of the sequence is $u_n=\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n\sqrt{5}}$ and $u_1=1,u_2=1$.

Answer to Problem 88AYU

Solution:

First $11$ terms are $1,1,2,3,5,8,13,21,34,55$ and $89$.

Explanation of Solution

Given information:

The Fibonacci sequence $u_{n+2}=u_{n+1}+u_n$ where the $n^{th}$ term of the sequence is $u_n=\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n\sqrt{5}}$ and $u_1=1,u_2=1$.

Explanation:

First two terms are given.

That is, $u_1=1,u_2=1$.

Now to find third term, substitute $n=1$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{1+2}=u_{1+1}+u_1$

$\Rightarrow u_3=u_2+u_1$

Substitute $u_1=1,u_2=1$ in above equation.

$\Rightarrow u_3=1+1=2$

To find fourth term substitute $n=2$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{2+2}=u_{2+1}+u_2$

$\Rightarrow u_4=u_3+u_2$

Substitute $u_2=1,u_3=2$ in above equation.

$\Rightarrow u_4=2+1=3$

To find fifth term substitute $n=3$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{3+2}=u_{3+1}+u_3$

$\Rightarrow u_5=u_4+u_3$

Substitute $u_3=2,u_4=3$ in above equation.

$\Rightarrow u_5=3+2=5$

To find sixth term substitute $n=4$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{4+2}=u_{4+1}+u_4$

$\Rightarrow u_6=u_5+u_4$

Substitute $u_4=3,u_5=5$ in above equation.

$\Rightarrow u_6=5+3=8$

To find seventh term substitute $n=5$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{5+2}=u_{5+1}+u_5$

$\Rightarrow u_7=u_6+u_5$

Substitute $u_5=5,u_6=8$ in above equation.

$\Rightarrow u_7=8+5=13$

To find eighth term substitute $n=6$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{6+2}=u_{6+1}+u_6$

$\Rightarrow u_8=u_7+u_6$

Substitute $u_6=8,u_7=13$ in above equation.

$\Rightarrow u_8=13+8=21$

To find ninth term substitute $n=7$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{7+2}=u_{7+1}+u_7$

$\Rightarrow u_9=u_8+u_7$

Substitute $u_7=13,u_8=21$ in above equation.

$\Rightarrow u_9=21+13=34$

To find tenth term substitute $n=8$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{8+2}=u_{8+1}+u_8$

$\Rightarrow u_{10}=u_9+u_8$

Substitute $u_8=21,u_9=34$ in above equation.

$\Rightarrow u_{10}=34+21=55$

To find eleventh term substitute $n=9$ in $u_{n+2}=u_{n+1}+u_n$.

$\Rightarrow u_{9+2}=u_{9+1}+u_9$

$\Rightarrow u_{11}=u_{10}+u_9$

Substitute $u_9=34,u_{10}=55$ in above equation.

$\Rightarrow u_{11}=55+34=89$

Therefore first $11$ terms are $1,1,2,3,5,8,13,21,34,55$ and $89$.

(b)

To determine

The first $10$ terms of the ratio $\frac{u_{n+1}}{u_n}$.

Answer to Problem 88AYU

Solution:

The first $10$ terms of the ratio $\frac{u_{n+1}}{u_n}$ are $1,\frac{2}{1},\frac{3}{2},\frac{5}{3},\frac{8}{5},\frac{13}{8},\frac{21}{13},\frac{34}{21},\frac{55}{34},\frac{89}{55}$.

Explanation of Solution

Given information:

The Fibonacci sequence $u_{n+2}=u_{n+1}+u_n$ where the $n^{th}$ term of the sequence is $u_n=\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n\sqrt{5}}$ and $u_1=1,u_2=1$

Explanation:

From part (a), $u_1=1,u_2=1,u_3=2,u_4=3,u_5=5,u_6=8,u_7=13,u_8=21,u_9=34,u_{10}=55$ and $u_{11}=89$.

To find first term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=1$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{1+1}}{u_1}=\frac{u_2}{u_1}$

Substitute $u_1=1,u_2=1$ in above equation.

Therefore, $\frac{u_2}{u_1}=\frac{1}{1}=1$.

To find second term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=2$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{2+1}}{u_2}=\frac{u_3}{u_2}$

Substitute $u_2=1,u_3=2$ in above equation.

Therefore, $\frac{u_3}{u_2}=\frac{2}{1}=2$.

To find third term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=3$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{3+1}}{u_3}=\frac{u_4}{u_3}$

Substitute $u_3=2,u_4=3$ in above equation.

Therefore, $\frac{u_4}{u_3}=\frac{3}{2}=1.5$

To find fourth term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=4$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{4+1}}{u_4}=\frac{u_5}{u_4}$

Substitute $u_4=3,u_5=5$ in above equation.

Therefore, $\frac{u_5}{u_4}=\frac{5}{3}\approx 1.667$.

To find fifth term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=5$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{5+1}}{u_5}=\frac{u_6}{u_5}$

Substitute $u_5=5,u_6=8$ in above equation.

Therefore, $\frac{u_6}{u_5}=\frac{8}{5}=1.6$.

To find sixth term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=6$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{6+1}}{u_6}=\frac{u_7}{u_6}$.

Substitute $u_6=8,u_7=13$ in above equation.

Therefore, $\frac{u_7}{u_6}=\frac{13}{8}=1.625$.

To find seventh term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=7$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{7+1}}{u_7}=\frac{u_8}{u_7}$.

Substitute $u_7=13,u_8=21$ in above equation.

Therefore, $\frac{u_8}{u_7}=\frac{21}{13}\approx 1.615$.

To find eighth term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=8$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{8+1}}{u_8}=\frac{u_9}{u_8}$.

Substitute $u_8=21,u_9=34$ in above equation.

Therefore, $\frac{u_9}{u_8}=\frac{34}{21}\approx 1.619$.

To find ninth term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=9$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{9+1}}{u_9}=\frac{u_{10}}{u_9}$.

Substitute $u_9=34,u_{10}=55$ in above equation.

Therefore, $\frac{u_{10}}{u_9}=\frac{55}{34}\approx 1.618$.

To find tenth term of ratio $\frac{u_{n+1}}{u_n}$, substitute $n=10$ in $\frac{u_{n+1}}{u_n}$.

$\Rightarrow \frac{u_{n+1}}{u_n}=\frac{u_{10+1}}{u_{10}}=\frac{u_{11}}{u_{10}}$.

Substitute $u_{10}=55,u_{11}=89$ in above equation.

Therefore, $\frac{u_{11}}{u_{10}}=\frac{89}{55}\approx 1.618$.

Thus, the first $10$ terms of the ratio $\frac{u_{n+1}}{u_n}$ are $1,\frac{2}{1},\frac{3}{2},\frac{5}{3},\frac{8}{5},\frac{13}{8},\frac{21}{13},\frac{34}{21},\frac{55}{34},\frac{89}{55}$.

(c)

To determine

The number that a ratio $\frac{u_{n+1}}{u_n}$ approach as $n$ gets larger.

Answer to Problem 88AYU

Solution:

The ratio approaches to $1.618$.

Explanation of Solution

Given information:

The Fibonacci sequence $u_{n+2}=u_{n+1}+u_n$ where the $n^{th}$ term of the sequence is $u_n=\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n\sqrt{5}}$ and $u_1=1,u_2=1$

Explanation:

By observing all fraction values in part (b) as $n$ gets larger, the ratio approaches to the number $1.618$.

The number $1.618$ is called as Golden ratio.

(d)

To determine

The first $10$ terms of the ratio $\frac{u_n}{u_n+1}$.

Answer to Problem 88AYU

Solution:

The first $10$ terms of the ratio $\frac{u_n}{u_n+1}$ are $1,\frac{2}{1},\frac{2}{3},\frac{3}{5},\frac{5}{8},\frac{8}{13},\frac{13}{21},\frac{21}{34},\frac{34}{55},\frac{55}{89}$.

Explanation of Solution

Given information:

The Fibonacci sequence $u_{n+2}=u_{n+1}+u_n$ where the $n^{th}$ term of the sequence is $u_n=\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n\sqrt{5}}$ and $u_1=1,u_2=1$

Explanation:

From part (a), $u_1=1,u_2=1,u_3=2,u_4=3,u_5=5,u_6=8,u_7=13,u_8=21,u_9=34,u_{10}=55$ and $u_{11}=89$.

To find first term of ratio $\frac{u_n}{u_n+1}$, substitute $n=1$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_1}{u_{1+1}}=\frac{u_1}{u_2}$

Substitute $u_1=1,u_2=1$ in above equation.

Therefore, $\frac{u_1}{u_2}=\frac{1}{1}=1$.

To find second term of ratio $\frac{u_n}{u_n+1}$, substitute $n=2$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_n+1}=\frac{u_2}{u_{2+1}}=\frac{u_2}{u_3}$

Substitute $u_2=1,u_3=2$ in above equation.

Therefore, $\frac{u_2}{u_3}=\frac{1}{2}=0.5$.

To find third term of ratio $\frac{u_n}{u_n+1}$, substitute $n=3$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_3}{u_{3+1}}=\frac{u_3}{u_4}$

Substitute $u_3=2,u_4=3$ in above equation.

Therefore, $\frac{u_3}{u_4}=\frac{2}{3}\approx 0.667$.

To find fourth term of ratio $\frac{u_n}{u_n+1}$, substitute $n=4$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_4}{u_{4+1}}=\frac{u_4}{u_5}$

Substitute $u_4=3,u_5=5$ in above equation.

Therefore, $\frac{u_4}{u_5}=\frac{3}{5}=0.6$.

To find fifth term of ratio $\frac{u_n}{u_n+1}$, substitute $n=5$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_5}{u_{5+1}}=\frac{u_5}{u_6}$

Substitute $u_5=5,u_6=8$ in above equation.

Therefore, $\frac{u_5}{u_6}=\frac{5}{8}=0.625$.

To find sixth term of ratio $\frac{u_n}{u_n+1}$, substitute $n=6$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_6}{u_{6+1}}=\frac{u_6}{u_7}$.

Substitute $u_6=8,u_7=13$ in above equation.

Therefore, $\frac{u_6}{u_7}=\frac{8}{13}\approx 0.615$.

To find seventh term of ratio $\frac{u_n}{u_n+1}$, substitute $n=7$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_7}{u_{7+1}}=\frac{u_7}{u_8}$.

Substitute $u_7=13,u_8=21$ in above equation.

Therefore, $\frac{u_7}{u_8}=\frac{13}{21}\approx 0.619$.

To find eighth term of ratio $\frac{u_n}{u_n+1}$, substitute $n=8$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_8}{u_{8+1}}=\frac{u_8}{u_9}$.

Substitute $u_8=21,u_9=34$ in above equation.

Therefore, $\frac{u_8}{u_9}=\frac{21}{34}\approx 0.618$.

To find ninth term of ratio $\frac{u_n}{u_n+1}$, substitute $n=9$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_9}{u_{9+1}}=\frac{u_9}{u_{10}}$.

Substitute $u_9=34,u_{10}=55$ in above equation.

Therefore, $\frac{u_9}{u_{10}}=\frac{34}{55}\approx 0.618$

To find tenth term of ratio $\frac{u_n}{u_n+1}$, substitute $n=10$ in $\frac{u_n}{u_n+1}$.

$\Rightarrow \frac{u_n}{u_{n+1}}=\frac{u_{10}}{u_{10+1}}=\frac{u_{10}}{u_{11}}$.

Substitute $u_{10}=55,u_{11}=89$ in above equation.

Therefore, $\frac{u_{10}}{u_{11}}=\frac{55}{89}\approx 0.618$.

Thus, the first $10$ terms of the ratio $\frac{u_n}{u_n+1}$ are $1,\frac{2}{1},\frac{2}{3},\frac{3}{5},\frac{5}{8},\frac{8}{13},\frac{13}{21},\frac{21}{34},\frac{34}{55},\frac{55}{89}$.

(e)

To determine

The number that a ratio $\frac{u_n}{u_{n+1}}$ approach as $n$ gets larger.

Answer to Problem 88AYU

Solution:

The ratio approaches to $0.618$.

Explanation of Solution

Given information:

The Fibonacci sequence $u_{n+2}=u_{n+1}+u_n$ where the $n^{th}$ term of the sequence is $u_n=\frac{{\left(1+\sqrt{5}\right)}^n-{\left(1-\sqrt{5}\right)}^n}{2^n\sqrt{5}}$ and $u_1=1,u_2=1$

Explanation:

By observing all fraction value in part (d) as $n$ gets larger, the ratio approaches to the number $0.618$.

The number $0.618$ is called as Conjugate golden ratio.