f(1)=3 , f(n)= 6*f(n-1) for n ≥ 2
In this problem, we are given a recursive formula for a function f(n), where f(1) is defined as 3 and f(n) is defined as 6 times f(n-1) for n greater than or equal to 2
In this problem, we are given a recursive formula for a function f(n), where f(1) is defined as 3 and f(n) is defined as 6 times f(n-1) for n greater than or equal to 2.
To find the value of f(n) for a given n, we can use the recursive formula to iteratively compute the results.
Let’s start by finding the value of f(2):
f(2) = 6 * f(2-1) = 6 * f(1) = 6 * 3 = 18
Next, we can find the value of f(3):
f(3) = 6 * f(3-1) = 6 * f(2) = 6 * 18 = 108
Continuing this process, we can find the values of f(4), f(5), and so on.
f(4) = 6 * f(4-1) = 6 * f(3) = 6 * 108 = 648
f(5) = 6 * f(5-1) = 6 * f(4) = 6 * 648 = 3888
We can see that the values of f(n) increase rapidly as n increases.
In general, the recursive formula f(n) = 6 * f(n-1) can be written as:
f(n) = 6^n * f(1)
Therefore, we can find the value of f(n) by multiplying 6 raised to the power of (n-1) with f(1).
To summarize:
f(1) = 3
f(n) = 6^n * f(1) for n greater than or equal to 2
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