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Question:

If a₁ = 1 and a_{n + 1} – 3a_n + 2 = 4n for every positive integer n, then a₁₀₀ equals

Explanation:

a₁ = 1

a_n+1 = 4n + 3a_n – 2

a₂ = 4 + 3(1) – 2 = 5 = 3² – 4

a₃ = 4(2) + 3(5) – 2 = 21 = 3³ – 6

a₄ = 4(3) + 3(21) – 2 = 73 = 3⁴ – 8

∴ a_n = 3ⁿ – 2(n)

∴ a₁₀₀ = 3¹⁰⁰ – 200

Hence, option (c).