Discussion

Explanation:

Given, f(1) = 1, f(n) = 3n - f(n - 1)

Substituting,
n = 2, f(2) = 3 × 2 - f(1) = 5
n = 3, f(3) = 3 × 3 - f(2) = 4
n = 4, f(4) = 3 × 4 - f(3) = 8
n = 5, f(5) = 3 × 5 - f(4) = 7
n = 6, f(6) = 3 × 6 - f(5) = 11
n = 7, f(7) = 3 × 7 - f(6) = 10

Here, f(1), f(3), f(5), and so on, i.e., f(odd number)  form an arithmetic progression with common difference of 3.

We need to find f(2023). Now, how many odd numbers are there from 1 till 2023 = (2023 + 1)/2 = 1012

∴ f(2023) = f(1) + (1012- 1) × 3
⇒ f(2023) = 1 + 1011 × 3 = 3034

Hence, 3034.

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