Math

Prove that for any natural n the value of the expression (4n+5)^2–9 is divisible by 4.

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  1. squaring (4 n + 5) ... 16 n^2 + 40 n + 25

    subtracting 9 ... 16 n^2 + 40 n + 16

    all the terms are divisible by 4

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  2. (4n+5)^2–9 = (4n+5)^2-3^2
    = (4n+5+3)(4n+5-3)
    = (4n+8)(4n-2)
    = 4(n+2)(4n-2)
    which is divisible by 4

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