Question

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

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

general term definition:
f(n) = \boxed{?}(\boxed{})^n

Explanation:

Step1: Analyze the sequence pattern

The sequence is \(-1, 1, -1, 1, \dots\). Notice that each term is the negative of the previous term. For a geometric sequence, the general term is \(f(n)=a\times r^n\), where \(a\) is the initial term and \(r\) is the common ratio.

Step2: Identify initial term and common ratio

The initial term \(f(0) = - 1\), so \(a=-1\). The common ratio \(r\) is \(\frac{f(n)}{f(n - 1)}=-1\) (from the recursive formula \(f(n)=-1\times f(n - 1)\)).

Step3: Write the general term

Substitute \(a = - 1\) and \(r=-1\) into the geometric sequence general term formula. So \(f(n)=-1\times(-1)^n\)? Wait, no, wait. Wait, when \(n = 0\), \(f(0)=-1\times(-1)^0=-1\times1=-1\), which matches. When \(n = 1\), \(f(1)=-1\times(-1)^1=-1\times(-1)=1\), which matches. Wait, but also, we can rewrite \(-1\times(-1)^n=(-1)^{n + 1}\), but the form given is \(f(n)=[\?]\times([\?])^n\). So the first blank is \(-1\) and the second blank is \(-1\), because \(f(n)=-1\times(-1)^n\). Let's check: for \(n = 0\), \(-1\times(-1)^0=-1\times1=-1\) (correct). For \(n = 1\), \(-1\times(-1)^1=-1\times(-1)=1\) (correct). For \(n = 2\), \(-1\times(-1)^2=-1\times1=-1\) (correct). Yes, that works.

Answer:

The general term is \(f(n)=\boldsymbol{-1}\times\boldsymbol{(-1)}^n\) (so the first box is \(-1\) and the second box is \(-1\)).