Let's analyze the given recurrence relation and find a formula for \(f(n)\).

The recurrence relation is given by:

\(f(n) = (n - 1) + f(n - 1)\)

This relation holds for \(n > 72\), but the structure suggests it applies for \(n > 2\) working down to the base case \(f(2)\). The base case is given as:

\(f(2) = 1\)

We can expand the recurrence relation step by step by substituting \(f(n-1)\), \(f(n-2)\), and so on, until we reach the base case \(f(2)\).

• For \(n\): \(f(n) = (n - 1) + f(n - 1)\)
• For \(n-1\): \(f(n - 1) = (n - 2) + f(n - 2)\)
• For \(n-2\): \(f(n - 2) = (n - 3) + f(n - 3)\)
• ...
• For \(3\): \(f(3) = (3 - 1) + f(2) = 2 + f(2)\)

Now, let's substitute these back into the expression for \(f(n)\):

\(f(n) = (n - 1) + f(n - 1)\)

\(f(n) = (n - 1) + [(n - 2) + f(n - 2)] = (n - 1) + (n - 2) + f(n - 2)\)

\(f(n) = (n - 1) + (n - 2) + [(n - 3) + f(n - 3)] = (n - 1) + (n - 2) + (n - 3) + f(n - 3)\)

Continuing this process, we will eventually reach \(f(2)\):

\(f(n) = (n - 1) + (n - 2) + (n - 3) + \dots + (3 - 1) + f(2)\)

\(f(n) = (n - 1) + (n - 2) + (n - 3) + \dots + 2 + f(2)\)

This is a sum of integers from 2 up to \(n-1\), plus \(f(2)\).

\(f(n) = \sum_{k=2}^{n-1} k + f(2)\)

The sum of the first \(m\) integers is given by the formula \(\frac{m(m+1)}{2}\). The sum of integers from 1 to \(n-1\) is \(\frac{(n-1)((n-1)+1)}{2} = \frac{(n-1)n}{2}\).

The sum from 2 to \(n-1\) can be calculated as the sum from 1 to \(n-1\) minus the first term (which is 1):

\(\sum_{k=2}^{n-1} k = \left(\sum_{k=1}^{n-1} k\right) - 1 = \frac{(n-1)n}{2} - 1\)

Now, substitute this back into the expression for \(f(n)\):

\(f(n) = \left(\frac{n(n-1)}{2} - 1\right) + f(2)\)

We are given \(f(2) = 1\).

\(f(n) = \left(\frac{n(n-1)}{2} - 1\right) + 1\)

\(f(n) = \frac{n(n-1)}{2}\)

This formula holds for \(n \ge 2\).

Let's verify for small values:

• \(f(2) = \frac{2(2-1)}{2} = \frac{2 \times 1}{2} = 1\). This matches the base case.
• Using the recurrence for \(n=3\): \(f(3) = (3-1) + f(2) = 2 + 1 = 3\). Using the formula: \(f(3) = \frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3\). This matches.
• Using the recurrence for \(n=4\): \(f(4) = (4-1) + f(3) = 3 + 3 = 6\). Using the formula: \(f(4) = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6\). This matches.

The derived formula \(f(n) = \frac{n(n-1)}{2}\) correctly represents the function defined by the recurrence relation and the base case.

Comparing this with the given options, the correct formula is \(\dfrac{n(n-1)}{2}\).

Revision Table: Solving Recurrence Relations

Additional Information: Understanding Recurrence Relation f(n)

A recurrence relation like \(f(n) = (n - 1) + f(n - 1)\) describes how a value at step \(n\) depends on the value at step \(n-1\) and possibly \(n\) itself. This specific structure, \(f(n) = g(n) + f(n-1)\), is a first-order linear non-homogeneous recurrence relation. The term \(g(n) = n-1\) is the non-homogeneous part. The homogeneous part is simple, just \(f(n) = f(n-1)\). Solving such recurrences often involves summing the \(g(k)\) terms from the base case up to \(n-1\).

In this problem, we essentially found that \(f(n)\) accumulates the terms \((n-1), (n-2), \dots, 2\) starting from the base value \(f(2)\). The sum of these terms, \((n-1) + (n-2) + \dots + 2\), is an arithmetic series. The sum of an arithmetic series \(a_1 + a_2 + \dots + a_m\) is \(\frac{m}{2}(a_1 + a_m)\). In our case, the series is \(2 + 3 + \dots + (n-1)\). Here, the terms are \(a_k = k\) starting from \(k=2\) up to \(n-1\). The number of terms is \((n-1) - 2 + 1 = n-2\). The first term is 2, and the last term is \(n-1\). The sum is \(\frac{n-2}{2}(2 + (n-1)) = \frac{n-2}{2}(n+1)\). This doesn't seem right, as we got \(\frac{n(n-1)}{2}-1\) from the sum from 1 to \(n-1\) minus 1.

Let's re-check the sum \(\sum_{k=2}^{n-1} k\). This is the sum of integers from 2 to \(n-1\). The standard formula for the sum of integers from 1 to \(m\) is \(\frac{m(m+1)}{2}\). Sum from 1 to \(n-1\) is \(\frac{(n-1)((n-1)+1)}{2} = \frac{(n-1)n}{2}\). The sum from 2 to \(n-1\) is the sum from 1 to \(n-1\) minus the term for \(k=1\): \(\sum_{k=2}^{n-1} k = \left(\sum_{k=1}^{n-1} k\right) - 1 = \frac{n(n-1)}{2} - 1\). This calculation was correct.

So, \(f(n) = \left(\frac{n(n-1)}{2} - 1\right) + f(2)\). With \(f(2)=1\), we get \(f(n) = \frac{n(n-1)}{2} - 1 + 1 = \frac{n(n-1)}{2}\).

The condition \(n > 72\) indicates the formula should be applied for values of \(n\) greater than 72. The derivation holds for \(n \ge 2\).