Question

1. a sequence is defined by (f(0)=4) and (f(n)=2cdot f(n - 1)). write the explicit rule for this function.
2. a sequence is defined by (f(0)= - 1) and (f(n)=f(n - 1)-3). write the explicit rule for this function.

Explanation:

Step1: Analyze the first - type sequence

For the sequence defined by \(f(0)=4\) and \(f(n) = 2\cdot f(n - 1)\), this is a geometric sequence with first - term \(a=f(0) = 4\) and common ratio \(r = 2\). The explicit formula for a geometric sequence is \(f(n)=a\cdot r^{n}\). Substituting \(a = 4\) and \(r = 2\) into the formula, we get \(f(n)=4\cdot2^{n}=2^{2}\cdot2^{n}=2^{n + 2}\).

Step2: Analyze the second - type sequence

For the sequence defined by \(f(0)=-1\) and \(f(n)=f(n - 1)-1\), this is an arithmetic sequence with first - term \(a=f(0)=-1\) and common difference \(d=-1\). The explicit formula for an arithmetic sequence is \(f(n)=a+(n)d\). Substituting \(a=-1\) and \(d = - 1\) into the formula, we get \(f(n)=-1+(n)\times(-1)=-1 - n\).

Answer:

1. The explicit rule for the sequence with \(f(0)=4\) and \(f(n)=2\cdot f(n - 1)\) is \(f(n)=2^{n + 2}\).
2. The explicit rule for the sequence with \(f(0)=-1\) and \(f(n)=f(n - 1)-1\) is \(f(n)=-n - 1\).