High School

Here is the initial term and recursive formula for a sequence:

[tex]
\[
\begin{align*}
f(1) & = 45 \\
f(n) & = f(n-1) + 1
\end{align*}
\]
[/tex]

Which explicit formula can be used to define this sequence?

A. [tex]\( f(n) = n + 46 \)[/tex]

B. [tex]\( f(n) = 45 \cdot 44^{n-1} \)[/tex]

C. [tex]\( f(n) = n + 44 \)[/tex]

D. [tex]\( f(n) = 44 \cdot 45^{n-1} \)[/tex]

Answer :

We start with the recursive definition:

[tex]$$
f(1)=45 \quad \text{and} \quad f(n) = f(n-1) + 1.
$$[/tex]

This tells us that the sequence is arithmetic with a common difference of 1. The general formula for an arithmetic sequence is:

[tex]$$
f(n) = a + (n-1)d
$$[/tex]

where [tex]$a$[/tex] is the first term and [tex]$d$[/tex] is the common difference. Here, we have:

- [tex]$a = 45$[/tex]
- [tex]$d = 1$[/tex]

Substitute these values into the formula:

[tex]$$
f(n) = 45 + (n-1) \cdot 1 = 45 + n - 1 = n + 44
$$[/tex]

Thus, the explicit formula for the sequence is:

[tex]$$
f(n) = n + 44.
$$[/tex]

Among the options provided, the correct explicit formula is:

[tex]$$
\boxed{f(n)=n+44.}
$$[/tex]

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