# 64|3^2n+2 - 8n - 9

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You need to perform mathematical induction to test if the given statement holds for all `n>=1` , hence, you need to start by showing that the statement holds for `n = 1` , such that:

`64|(3^(2+2) - 8 - 9) => 64|(81 - 17) => 64|64`

Since the basis step proves that the statement `P(n)` holds for `n = 1` , you need to perform the second step, called inductive step. You need to prove that if `P(k)` holds, then `P(k+1)` holds, such that:

`P(k): 64|(3^(2k+2) - 8k - 9)` holds

`P(k): 64|(9^(k+1) - 8k - 9) => P(k): 64|(9(9^k - 1) - 8k)`

`For k =1 => P(1): 64|(9*8 - 8k) => P(1): 64|8*(9 - k)`

You need to prove that `P(k+1)` also holds, such that:

`P(k+1): 64|(3^(2(k+1)+2) - 8(k+1) - 9)`

`P(k+1): 64|(9^(k+2) - 8(k+1) - 9)`

`P(k+1): 64|(9*9^(k+1) - 8k - 8 - 9)`

`P(k+1): 64|(9*(9^(k+1) - 1) - 8k - 8)`

`P(k+1): 64|(9*(9^(k+1) - 1) - 8(k+1))` holds

**Hence, since the two steps of mathematical induction hold, then the statement **`P(n): 64|(3^(2n+2) - 8n - 9) ` holds for all `n>=1.`

`holds for all n>=1.`

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