Question

Find a possible function f(n) whose domain is a set of natural numbers and whose outputs are the terms of the sequence. -4^1, 4^(-1/2), 1, 4^(1/2), 4

Answer 1

we have the sequence

4^=-1, 4^(-1/2), 1, 4^(1/2), 4​

so

a1=4^-1=1/14

a2=4^(-1/2)=(1/4)^(1/2)=1/2

a3=1

a4=4^(1/2)=2

a5=4

therefore

a2/a1=(1/2)/(1/4)=2

a3/a2=1/(1/2)=2

a4/a3=2/1=2

a5/a4=4/2=2

that means

Is a geometric sequence and the common ratio is r=2

the general formula is equal to

$f(n)=a1\cdot(r)^{(n-1)}$

substitute given values

a1=1/4

r=2

$f(n)=\frac{1}{4}\cdot(2)^{(n-1)}$

we have the function f(n)

Verify the outputs

For n=1

substitute

$\begin{gathered} f(1)=\frac{1}{4}\cdot(2)^{(1-1)} \\ f(1)=\frac{1}{4}\cdot(2)^{(0)} \\ f(1)=\frac{1}{4}=4^{-1} \end{gathered}$

For n=2

$\begin{gathered} f(2)=\frac{1}{4}\cdot(2)^{(2-1)} \\ f(2)=\frac{1}{4}\cdot(2)^{(1)} \\ f(2)=\frac{1}{2} \end{gathered}$

the general expression is

$f(n)=\frac{1}{4}\cdot(2)^{(n-1)}$

Remmeber that

(1/4)=4^-1=(2^2)^-1=2^-2

substitute in the given expression

$\begin{gathered} f(n)=\frac{1}{4}\cdot(2)^{(n-1)} \\ f(n)=(2)^{(-2)}\cdot(2)^{(n-1)} \end{gathered}$

Adds the exponents

-2+(n-1)=n-3

therefore

$f(n)=2^{(n-3)}$

an equivalent expression