Chapter 7.5, Problem 35E

Solution

To state: The recursive rule for the sequence $1,2,12,56,272\cdots$ .

The resultant recursive rule is $a_n=4\left(a_{n-2}+a_{n-1}\right)$ .

Given information:

The given sequence is $1,2,12,56,272\cdots$ .

Explanation:

Consider the given sequence,

  $1,2,12,56,272\cdots$

The first term of the sequence is: $a_1=1$

The common difference ( d ) is:

  $\begin{array}{c}2-1=1\\ 12-2=10\end{array}$

The given sequence is not an arithmetic sequence because the common difference of two successive terms is not equal.

Now check if the sequence is geometric.

The common ratio ( r ) is:

  $\begin{array}{c}r=\frac{2}{1}\\ =2\\ r=\frac{12}{2}\\ =6\end{array}$

The given sequence is not a geometric sequence because the common ratio is not same from the two terms.

Now again consider the sequence $1,2,12,56,272\cdots$ ,

Beginning with the third term in the sequence, $a_3$ can be obtained by multiplying the sum of first term and the second term by 4.

  $\begin{array}{c}{a}_3=4\left(a_1+a_2\right)\\ =4\left(1+2\right)\\ =12\end{array}$

Similarly, the fourth term can be obtained by multiplying the sum of third term and the second term by 4.

  $\begin{array}{c}{a}_4=4\left(a_3+a_2\right)\\ =4\left(12+2\right)\\ =56\end{array}$

So the recursive rule for the sequence $a_1=1,a_2=2$ is $a_n=4\left(a_{n-2}+a_{n-1}\right)$ .