# Thread: mathematical induction problem

1. ## mathematical induction problem

Use mathematical induction to prove the following:
a factor of (9^n - 8n - 1) is 64 for all n > or = 2

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

where a = any integer

9^k - 8k - 1 = 64a

prove that 9^(k+1) - 8(k+1) - 1 = 64b; where b = any integer
9(9^k) - 8k - 9

so that's how far i've gotten and now i'm stuck. please help me.

2. First show that the statement is true for n = 2. Which infact is.

Now assume

$\displaystyle 9^k - 8k - 1$ is true.

Consider

$\displaystyle 9^ (k + 1) - 8(k + 1) - 1$

Which is same as
$\displaystyle 9 * 9^k - 8k - 9$

which on simplification gives $\displaystyle 9 (9^k - 8k - 1) - 64k$

Now by assumption $\displaystyle 9 (9^k - 8k - 1)$is divisible by 64 and for k > 2 64k is divisible by 64.
So statement is true for k + 1

Hence proved.

P.S I am new to this forum and am learning how to type mathematical expressions. So ignore any typing mistake.

3. Originally Posted by oblixps
Use mathematical induction to prove the following:
a factor of (9^n - 8n - 1) is 64 for all n > or = 2

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

where a = any integer

9^k - 8k - 1 = 64a

prove that 9^(k+1) - 8(k+1) - 1 = 64b; where b = any integer

9(9^k) - 8k - 9
$\displaystyle 9(9^k) - 9 - 8k$

$\displaystyle 9(9^k - 1) - 8k$

$\displaystyle 9(9^k - 8k - 1) - 8k + 9(8k)$

$\displaystyle 9(9^k - 8k - 1) + 8(8k)$

$\displaystyle 9(9^k - 8k - 1) + 64k$

$\displaystyle 9(64a) + 64k$

$\displaystyle 64(9a + k)$

4. Thanks skeeter. That was very clear way to explain. I went bit gross in my explanation.

5. thank you very much for the help guys!

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### prove that 64 is a factor of 9^n-8n-1

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