Let a0 = 0 and ax = 3an–1 + 1 for n ≥ 1. Then the remainder obtain

1.	Let a0 = 0 and ax = 3an–1 + 1 for n ≥ 1. Then the remainder obtained dividing a2010 by 11 is- (A) 0 (B) 7 (C) 3 (D) 4
Answer» Correct Option :- (A) 0 Explanation :- a_n= 3a_n–1 + 1 a₂₀₁₀ = 3a₂₀₀₉ + 1 = 3(3a₂₀₀₈ + 1) + 1 = 3² a₂₀₀₈ + 3 + 1 = 3³ a₂₀₀₇ + 3 + 3 + 1 . . . . 3²⁰¹⁰ a₀ +(3 + 3 + ...3) + 1 ------{2009 times} = 0 + 6027 + 1 = 6028 Remainder (6028/11) = 0

Let a0 = 0 and ax = 3an–1 + 1 for n ≥ 1. Then the remainder obtained dividing a2010 by 11 is- (A) 0 (B) 7 (C) 3 (D) 4

Answer»

Correct Option :- (A) 0

Explanation :-

a_n= 3a_n–1 + 1

a₂₀₁₀ = 3a₂₀₀₉ + 1

= 3(3a₂₀₀₈ + 1) + 1

= 3² a₂₀₀₈ + 3 + 1

= 3³ a₂₀₀₇ + 3 + 3 + 1

3²⁰¹⁰ a₀ +(3 + 3 + ...3) + 1 ------{2009 times}

= 0 + 6027 + 1 = 6028

Remainder (6028/11) = 0

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Let a0 = 0 and ax = 3an–1 + 1 for n ≥ 1. Then the remainder obtained dividing a2010 by 11 is- (A) 0 (B) 7 (C) 3 (D) 4

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