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prove that for any positive integer n, the value of 3^2n+2 - 8n-9 is divisible by 64

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  1. The way you typed it, 3^2n+2 - 8n-9
    it isn't even true for n = 1

    You must have meant:
    3^(2n+2) - 8n-9

    let's use induction:
    step 1. is it true for n = 1 ?
    3(4) - 8(1) - 9 = 64, which is clearly divisible by 64
    step 2. Assume it is true for n = k
    i.e. 3^(2k+2) - 8k - 9 is divisible by 64

    step 3. Is 3^(2(k+1) + 2) - 8(k+1) - 9 divisible by 64

    it is if the difference between the expressions in step 3 and step 2 is divisible by 64. Let's take that difference.
    (concept: the difference between any two multiples of p must be divisible by p)
    e.g. 72 and 192 are both multiples of 12, then 192-72 = 120, which is a multiple of 12

    3^(2(k+1) + 2) - 8(k+1) - 9 - (3^(2k+2) - 8k - 9)
    = 3^(2k + 4) - 8k - 8 - 9 - 3^(2k+2) + 8k + 9
    = 3^(2k+4) - 3^(2k+2) - 8
    = 3^(2k+2)( 3^2 - 1) - 8
    = 3^(2k+2)(8) - 8
    = 3^(2k)(3^2)(8) - 8
    = 72(3^(2k)) - 8
    = 72(3^2)^k - 8
    = 72(9^k) - 8
    = 8(9^(k+1)) - 8
    = 8( 9^(k+1) - 1) ****
    argghhh, now I have to show that 9^(k+1) - 1 is divisible by 8
    step 1 , works for k=1
    step 2, assum 9^(k+1) - 1 is divisible by 8
    step 3, what about 9^(k+2) - 1 ???

    again take the difference .....
    9^(k+2) - 1 - (9^(k+1) - 1)
    = 9^(k+2) - 9^(k+1)
    = 9^(k+1)( 9^1 - 1)
    = 9^(k+1)(8), which means 9^(k+2) - 1 is divisible by 8, Yeahhh

    so in **** we have 8(multiple of 8) = a multiple of 64

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