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    prove that for all natural values of n, 64 | [(9^n)-8n-1]

    i got to 9^(n+1) - 8(n+1) -1 = 64k + 8(9^n) -8

    but taking a factor of 64 out leaves u with eighths inside the bracket-does this prove the statement in bold?? or have i just made a total mess of it?!

    any help would be much appreciated
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    I don't understand what you have to prove? 64 | [(9^n)-8n-1] ??

    What does this mean?
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    that 64 divides the expression
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    (Original post by toad)
    prove that for all natural values of n, 64 | [(9^n)-8n-1]

    i got to 9^(n+1) - 8(n+1) -1 = 64k + 8(9^n) -8

    but taking a factor of 64 out leaves u with eighths inside the bracket-does this prove the statement in bold?? or have i just made a total mess of it?!

    any help would be much appreciated
    8(9^n)-8 = 8(9^n -1 ), now 9^n -1 is divisible by 8 for all n (factorise), so you can write it as 64m for some m.
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    [9^(n + 1) - 8(n + 1) - 1] - [9^n - 8n - 1]
    = 8*9^n - 8
    = 8((9^n) - 1)
    = 8(9 - 1)[9^(n - 1) + 9^(n - 2) + ... + 9 + 1]
    = 64 [9^(n - 1) + 9^(n - 2) + ... + 9 + 1]

    is divisible by 64. The result follows by induction.
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    sorry how did u get from this line = 8((9^n) - 1)

    to this one = 8(9 - 1)[9^(n - 1) + 9^(n - 2) + ... + 9 + 1] ??
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    oh rite i get u! sorry never seen that before
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    9^(n+1) - 8(n+1) -1 = 9(9^n - 8n -1) - 64n
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    prove that for all natural values of n, 64 | [(9^n)-8n-1]

    Here's what I got then.

    Works for n = 1:

    9^1 - 8*1 - 1 = 0
    this divides 64

    Now assume it works for all natural values of n:

    64 divides (9^n - 8n - 1)

    That is, for any positive integer a,

    64a = 9^n - 8n - 1
    9^n = 64a + 8n + 1

    Now the proof that if it is true for one natural number n, it is also true for n+1 - that:

    64 also divides (9^(n+1) - 8(n+1) - 1)

    9^(n+1) - 8(n+1) - 1 = 9*9^n - 8n - 9 = (64a + 8n + 1) - 8n - 1 = 64a

    Shown.
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    To prove this without induction consider the binomial expansion of (8+1)^n. Except for the last two terms the power of 8 in each term is at least two. That is,

    (8+1)^n = 8^n + (nC1)8^(n-1) + (nC2)8^(n-2) + ... + (nC3)8^3 + (nC2)8^2 + (nC1)8 + 1
    = K(8^2) + 8n + 1

    where (nCr) represents the binomial coefficients and K is some integer. This result rearranges to give

    9^n - 8n - 1 = 64K

    which is what we wanted.
 
 
 
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