Chapter 8.1, Problem 62E

To determine

To write: the first five terms of the sequence defined recursively and write the pattern for nth term.

Answer to Problem 62E

The first five terms are

  $14,-28,56,-112,224$ .

The pattern for nth term is $a_n={\left(-1\right)}^{n+1}\times 14\times 2^{n-1}$ .

Explanation of Solution

Given information:

Given sequence

  $a_1=14,a_{k+1}=-2a_k$

Calculation:

Since the first term of the sequence is $a_1=14$ therefore, the subsequent terms can be computed from the recursive formula, $a_{k+1}=-2a_k$ as follows,

  $\begin{array}{c}{a}_2=-2a_1\\ =-2\times 14\\ =-28\\ a_3=-2a_2\\ =-2\times \left(-28\right)\\ =56\\ a_4=-2a_3\\ =\left(-2\right)56\\ =-112\\ a_5=-2a_4\\ =-2\times \left(-112\right)\\ =224\end{array}$

Hence the first five terms are

  $14,-28,56,-112,224$ .

Now, the pattern for nth term can be obtained as

Consider the sequence $14,-28,56,-112,224$

For $n=1,2,3,4,5,\mathrm{...}n$

The terms are $14,-28,56,-112,\mathrm{224...}{a}_n$

Rewrite the terms as shown below,

  $\begin{array}{c}14,-28,56,-112,\mathrm{224...}{a}_n=14\times 2^0,{\left(-1\right)}^{2+1}\times 14\times 2^{2-1},{\left(-1\right)}^{3+1}\times 14\times 2^{3-1},{\left(-1\right)}^{3+1}\times 14\times 2^{4-1},\mathrm{..},a_n\\ ={\left(-1\right)}^{n+1}\times 14\times 2^{n-1}\end{array}$

The pattern followed is such that each term at even position is negative and every term is multiplied by 14 times 2 raise to the power n minus 1.

. Therefore , the general term is $a_n={\left(-1\right)}^{n+1}\times 14\times 2^{n-1}$ .