Chapter 7, Problem 36REAE

Solution

To state: The recursive rule for the sequence $4,6,9,13,18,\cdots$ .

The resultant recursive rule is $a_n=n+a_{n-1}$ .

Given information:

The given sequence is $4,6,9,13,18,\cdots$ .

Explanation:

Consider the given sequence,

  $4,6,9,13,18,\cdots$

The first term of the sequence is: $a_1=4$

The common difference ( d ) is:

  $\begin{array}{c}6-4=2\\ 9-6=3\\ 13-9=4\end{array}$

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:

  $\begin{array}{c}r=\frac{6}{4},r=\frac{3}{2}\\ r=\frac{9}{6},r=\frac{3}{2}\\ r=\frac{13}{9}\end{array}$

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

Now again consider the sequence $4,6,9,13,18,\cdots$ ,

Beginning with the second term in the sequence, $a_2$ can be obtained by adding the previous term and n .

Similarly, the third term can be obtained by adding the previous term and n .

So the recursive rule for the sequence $a_1=4,a_2=6$ is $a_n=n+a_{n-1}$ .