Question

which recursive formula can be used to generate the sequence shown, where $f(1) = 5$ and $n \geq 1$?
$5, -1, -7, -13, -19, \dots$
\bigcirc $f(n + 1) = f(n) + 6$
\bigcirc $f(n) = f(n + 1) - 6$
\bigcirc $f(n + 1) = f(n) - 6$
\bigcirc $f(n + 1) = -6f(n)$

Explanation:

Step1: Analyze the sequence difference

The sequence is \(5, -1, -7, -13, -19, \dots\). Let's find the difference between consecutive terms.
\(-1 - 5=-6\), \(-7 - (-1)=-6\), \(-13 - (-7)=-6\), \(-19 - (-13)=-6\). So each term is obtained by subtracting 6 from the previous term.

Step2: Match with recursive formulas

A recursive formula for a sequence defines the \(n + 1\)-th term in terms of the \(n\)-th term. Since each term \(f(n + 1)\) is \(f(n)-6\) (because we subtract 6 from the previous term to get the next term), we check the options.

• Option 1: \(f(n + 1)=f(n)+6\) would give positive differences, which is wrong.
• Option 2: \(f(n)=f(n + 1)-6\) can be rearranged to \(f(n + 1)=f(n)+6\), same as option 1, wrong.
• Option 3: \(f(n + 1)=f(n)-6\) matches our analysis.
• Option 4: \(f(n + 1)=-6f(n)\) would give a geometric sequence (multiplicative), but our sequence is arithmetic (additive difference), so wrong.

Answer:

C. \(f(n + 1)=f(n)-6\) (where C is the third option in the list: \(f(n + 1)=f(n)-6\))