Given the recurrence relation f(n) = (n-1) + f(n-1), n > 72, f(2) = 1, then f(n) is:
Question
Given the recurrence relation
f(n) = (n-1) + f(n-1), n > 72, f(2) = 1, then f(n) is:
Solution
The given recurrence relation is f(n) = (n-1) + f(n-1), n > 72, f(2) = 1.
To solve this, we can start by expanding the recurrence relation:
f(n) = (n-1) + f(n-1) = (n-1) + ((n-2) + f(n-2)) = (n-1) + (n-2) + ((n-3) + f(n-3)) = (n-1) + (n-2) + (n-3) + ... + 1 + f(2)
Since f(2) = 1, we can substitute it into the equation:
f(n) = (n-1) + (n-2) + (n-3) + ... + 1 + 1 = 1 + 2 + 3 + ... + (n-1) + 1 = n + (1 + 2 + 3 + ... + (n-1))
The sum of the first (n-1) natural numbers is given by the formula n(n-1)/2. So we can substitute this into the equation:
f(n) = n + n(n-1)/2 = n/2 * (n+1)
So, the solution to the recurrence relation is f(n) = n/2 * (n+1) for n > 72.
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