Discussion

Explanation:

Given, f(n) = f(f(n – 1)) + f(n – f(n – 1)),

f(1) = 1 and f(2) = 1

putting n = 3, we get

f(3) = f(f(3 – 1)) + f(3 – f(3 – 1))

⇒ f(3) = f(f(2)) + f(3 – f(2))

⇒ f(3) = f(1) + f(3 – 1)

⇒ f(3) = 1 + f(2)

⇒ f(3) = 1 + 1 = 2

Now, putting n = 4, we get

f(4) = f(f(4 – 1)) + f(4 – f(4 – 1))

⇒ f(4) = f(f(3)) + f(4 – f(3))

⇒ f(4) = f(2) + f(4 – 2)

⇒ f(4) = 1 + f(2)

⇒ f(4) = 1 + 1 = 2

Hence, option (b).

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