Chapter 9.1, Problem 57E

To determine

Find the first 12 terms of the Fibonacci sequence whose 10^th term is given.

Answer to Problem 57E

The terms are

1,1,2,3,5,8,13,21,34,55,89,144,233

Explanation of Solution

Given:

The 10^thterm

  $b_n=\frac{a_2+1}{a_2}n\ge 1$

The Fibonacci sequence is defined recursively as,

  $a_0=1,a_1=1,a_2=a_{2-2}+a_{2-1}$

Here, $k\ge 2$

Therefore, the first 12 terms of Fibonacci sequence are,

  $\begin{array}{l}{a}_0=1\hfill \\ a_1=1\hfill \end{array}$

  $\begin{array}{cc}{a}_2& =a_0+a_1\\ & =1+1\\ & =2\\ & \end{array}$

  $\begin{array}{cc}{a}_3& =a_1+a_2\\ & =1+2\\ & =3\end{array}$

  $\begin{array}{l}{a}_4=a_2+a_3\\ \begin{array}{l}=2+3\hfill \\ =5\hfill \end{array}\end{array}$

  $\begin{array}{l}{a}_5=a_3+a_4\\ \begin{array}{l}=3+5\hfill \\ =8\hfill \end{array}\end{array}$

  $\begin{array}{l}{a}_6=a_4+a_5\hfill \\ =5+8\hfill \\ =13\hfill \end{array}$

  $\begin{array}{l}{a}_7=a_5+a_6\hfill \\ =8+13\hfill \\ =21\hfill \end{array}$

  $\begin{array}{l}{a}_8=a_6+a_7\\ =13+21 \\ =34\end{array}$

  $\begin{array}{cc}{a}_9& =a_7+a_8\\ & =21+34\\ & =55\\ a_{10}& =a_8+a_9\\ & =34+55\\ & =89\\ a_{11}& =a_9+a_{10}\\ & =55+89\\ & =144\\ a_{12}& =a_{10}+a_{11}\\ & =89+144\\ & =233\end{array}$

Thus the terms are

1,1,2,3,5,8,13,21,34,55,89,144,233