# Write the first five terms of the sequence defined recursively. Use the pattern to write the nth term of the sequence as a function of n. `a_1 = 14` `a_(k+1) = (-2)a_k`

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### 2 Answers

Since the first term is already known, to solve for the next four terms, starts with k=1.

`k=1`

`a_(1+1)=-2a_1`

`a_2=-2(14)`

`a_2=-28`

`k=2`

`a_(2+1)=-2a_2`

`a_3=-2(-28)`

`a_3=56`

`k=3`

`a_(3+1)=-2a_3`

`a_4=-2(56)`

`a_4=-112`

`k=4`

`a_(4+1)=-2a_4`

`a_5=-2(-112)`

`a_5=224`

**Thus, the first five terms of the sequence are 14,-28,56,-112 and 224.**

To solve for the nth term without using the recursive formula, we have to take note that when solving for a2, a3, a4 and a5, the factor -2 is always present. So, if we express the terms above as factors of -2 , it becomes:

`a_1=(-2)(-7)`

`a_2=-28=(-2)(-2)(-7)`

`a_3=56=(-2)(-2)(-2)(-7)`

`a_4=-112=(-2)(-2)(-2)(-2)(-7)`

`a_5=224=(-2)(-2)(-2)(-2)(-2)(-7)`

Notice that the value of k determines the number of times that -2 is multiplied by itself. Also, the factor -7 is always present in our terms. Moreover, only one factor of -7 is present in each term.

**Thus, without using the recursive formula, to get the nth term we may apply the formula `a_k=-7*(-2)^n` .**

### User Comments

The first term of a sequence that can be defined recursively is `a_1 = 14` and the (k+1)th term can be written in terms of the `k_(th)` term as `a_(k+1) = -2*(a_k)`

The second term of the sequence is the product of -2 and the first term:

`a_2 = -2*a_1 = -2*14 = -28`

Similarly, the third term is the product of -2 and the second term:

`a_3 = -2*a_2 = -2*-28 = 56`

The fourth term is:

`a_4 = -2*a_3 = -2*56 = -112`

and `a_5 = -2*-112 = 224`

If the terms are to be written as a function of n, start with `t_n`

`14 = 14*2^(n-1)*(-1)^(n+1)`

Using the same rule `t_2 = 14*2^(1)*(-1)^3 = -28`

`t_3 = 14*2^(2)*(-1)^(4) = 56`

This shows that as a function of n, the terms of the sequence can be written as `f(n) = 14*2^(n-1)*(-1)^(n+1)`