Question

\begin{cases}
f(1)=-62
\\\\
f(n)=f(n-1)+5
\end{cases}Find an explicit formula for
\[f(n)\].

Answer 1

To find an explicit formula for the function \( f(n) \) defined by the given recurrence relation, we begin by examining the information provided:

1. The initial value:
\[
f(1) = -62
\]

2. The recurrence relation:
\[
f(n) = f(n-1) + 5
\]

We can use this recurrence relation to generate a few more values:

- For \( n = 2 \):
\[
f(2) = f(1) + 5 = -62 + 5 = -57
\]

- For \( n = 3 \):
\[
f(3) = f(2) + 5 = -57 + 5 = -52
\]

- For \( n = 4 \):
\[
f(4) = f(3) + 5 = -52 + 5 = -47
\]

From these calculations, we can observe a pattern emerging. Specifically, each increment in \( n \) results in an increase of 5 in \( f(n) \) from the previous value. Therefore, we can express \( f(n) \) in terms of \( n \) and its initial condition by recognizing that every step from \( n = 1 \) to \( n \) includes \( (n-1) \) increments of 5.

Thus, we can express \( f(n) \) generally as:
\[
f(n) = f(1) + 5 \times (n-1)
\]

Substituting \( f(1) = -62 \):
\[
f(n) = -62 + 5(n - 1)
\]

We can simplify this expression:
\[
f(n) = -62 + 5n - 5 = 5n - 67
\]

Therefore, we have found the explicit formula for \( f(n) \):
\[
\boxed{5n - 67}
\]