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64. If A = 111111111 , prove that An = 3n-13n-13n-13n-13n-13n-13n-13n-13n-1 , n ∈ N.

0 3 Views | Posted 4 months ago
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  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    We have,

    (E) P (n) : If A = [
    111111111     ]. then An = [3n−13n−13n−13n−13n−13n−13n−13n−13n−1] n ? N.

    P (1) : A1= [31−131−131−131−131−131−131−131−131−1] = [303030303030303030] =[
    111111111     ]

    So, the result holds true for n = 1.

    Let the result be for n = k. So,

    P (k): Au= [3k−13k−13k−13k−13k−13k−13k−13k−13k−1]

    Then P (k+1): Ak + 1 = Ak. A = [3k−13k−13k−13k−13k−13k−13k−13k−13k−1] [
    111111111 ]

    = [3k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−1]

    = Ak + 1 = [3(k+1)−13(k+1)−13(k+1)−13(k+1)−13(k+1)−13(k+1)−13(k+1)−13(k+1)−13(k+1)−1]

    The result holds for n = k + 1. Hence,

    An = [3n−13n−13n−13n−13n−13n−13n−13n−13n−1] holds for all natural number.

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