Question

Write the recursive and explicit functions the following geometric sequence.2, 4, 8, 16.

Answer 1

We were given the following numbers:

$2,4,8,16$

We are to write the recursive and explicit functions for this geometric function.

A recursive function is one that requires that the previous term be known to obtain the successive term in the sequence

An explicit function is one that defines the value of a term based on its current position in the sequence

This is shown below:

$\begin{gathered} a_1=2 \\ a_2=4 \\ a_3=8 \\ a_4=16 \\ r=\frac{a_2}{a_1}\equiv\frac{a_3}{a_2}\equiv\frac{a_4}{a_3}\equiv\frac{4}{2}\equiv\frac{8}{4}\equiv\frac{16}{8}=2 \\ r=2 \end{gathered}$

For the Recursive function, we have:

$\begin{gathered} a_n=r\cdot a_{n-1} \\ a_1=2 \\ r=2 \\ \Rightarrow a_n=2\cdot a_{n-1} \\ \text{Let's check if this is true:} \\ when\colon n=2 \\ a_2=2\cdot a_{2-1}\Rightarrow a_2=2\cdot a_1\Rightarrow a_2=2\cdot2=4 \\ when\colon n=3 \\ a_3=2\cdot a_{3-1}\Rightarrow a_3=2\cdot a_2\Rightarrow a_3=2\cdot4=8 \\ \\ \therefore a_n=2\cdot a_{n-1} \end{gathered}$

For the Explicit function, we have:

$\begin{gathered} a_n=a_1\cdot r^{n-1} \\ r=2 \\ a_1=2\cdot2^{1-1}=2\times1=2 \\ a_2=2\cdot2^{2-1}=2\times2=4^{} \\ a_3=2\times2^{3-1}^{}=2\times2^2=8 \\ a_4=2\times2^{4-1}=2\times2^3=16 \\ \\ \therefore a_n=a_1\cdot r^{n-1} \end{gathered}$