Question

finally, what is the general term definition for this sequence?
1, 2, 4, 8, ...
recursive definition:
\

$$\begin{cases} f(0) = 1 \\\\ f(n) = 2f(n - 1) \\end{cases}$$

general term definition:
$f(n) = ?(\quad)^n$

Explanation:

Step1: Identify the sequence type

The sequence \(1, 2, 4, 8, \dots\) is a geometric sequence. In a geometric sequence, each term is obtained by multiplying the previous term by a common ratio \(r\). Here, the common ratio \(r = \frac{2}{1}=\frac{4}{2}=\frac{8}{4}=2\).

Step2: Recall the formula for the general term of a geometric sequence

The general term of a geometric sequence with first term \(a\) (when \(n = 0\)) and common ratio \(r\) is given by \(f(n)=a\times r^{n}\).

Step3: Determine the values of \(a\) and \(r\)

From the recursive definition, when \(n = 0\), \(f(0)=1\), so the first term \(a = 1\). The common ratio \(r = 2\) (as we found in Step 1).

Step4: Substitute \(a\) and \(r\) into the formula

Substituting \(a = 1\) and \(r = 2\) into the formula \(f(n)=a\times r^{n}\), we get \(f(n)=1\times(2)^{n}\), which simplifies to \(f(n)=1\times(2)^{n}\) or \(f(n) = 2^{n}\) (since \(1\times x=x\) for any \(x\)).

Answer:

The first box (the coefficient) is \(1\) and the second box (the base of the exponent) is \(2\), so \(f(n)=\boldsymbol{1}\times(\boldsymbol{2})^{n}\)