I think I have this one right. Can anyone help confirm if I've solved this correctly?

Many thanks.

Q.13^{n}- 6^{n-2}is divisible by 7 for n $\displaystyle \in \mathbb{N}_0$

Attempt:Step 1:For n = 1,

13^{1 }- 6^{1-2 }= 13 - 6^{-1}= 13 - 0.166 = 12.833, which cannot be divided by 7.

However, if n = n + 1: 13^{n}- 6^{n-2 }= 13^{n+1}- 6^{n-1}

Now, for n = 1,

13^{2}- 6^{0}= 169 - 1 = 168, which can be divided by 7.

Assume the statement is true for n = k,

Step 2:

i.e. assume 13^{k+1}- 6^{k-1 }can be divided by 7

13^{k+1 }- 6^{k-1}= 7Z, where Z is an integer...1

We must now show that the statement is true for n = k + 1,

i.e. 13^{k+2}- 6 can be divided by 7.

13^{k+}^{2}- 13(6^{k}^{-1}) + 13(6^{k-1}) - 6^{k}= 13(13^{k+1}- 6^{k-1}) - 6^{k-1}(6 - 13)...from1above = 13(7Z) - (-7)(6^{k-1}) = 91Z + 7(6^{k-1}) = 7[13Z + 6^{k-1}], which can be divided by 7.

Therefore, the statement is true for n = k + 1, assuming it is true for n = k.

Thus the statement is true for n = 2, 3... & all n $\displaystyle \in \mathbb{N}_0$