Question

How to prove that `7*25^n+2*6^(n+1)` divides 19?

Answer 1

Induction Proof - Hypothesis

We seek to prove that the expression:

`f(n) = 7*25^n+2*6^(n+1)` ..... [A]

is divisible by `19 AA n in NN` So let us test this assertion using Mathematical Induction:

Induction Proof - Base case:

When `n=1` the given expression gives:

`f(1) = 7*25+2*6^2 = 175+72=247=13*19`

Which is divisible by `19`, So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`f(m) = 7*25^m+2*6^(m+1) = 19k` ..... [B]

Where `k in NN`. Now, consider the expression [A] when we have `n=m+1`:

`f(m+1) = 7*25^(m+1)+2*6^(m+2)`
`" " = 7*25^(m+1)+2*6^((m+1)+1)`
`" " = 7*25^(m)*25+2*6^(m+1)*6`
`" " = 7*25^(m)*(19+6)+2*6^(m+1)*(6)`

`" " = 7*25^(m)*6+7*25^(m)*19+2*6^(m+1)*6`

`" " = 6*(7*25^(m)+2*6^(m+1))+19*7*25^(m)`
`" " = 6*(19k)+19*7*25^(m) \ \ \` using [B]
`" " = 19*(6k+7*25^(m) )`

Which is also divisible by `19`

Induction Proof - Summary

So, we have shown that if the given result [A] is true for `n=m`, then it is also true for `n=m+1` where `m gt 1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for `n in NN`

Hence we have:

`f(n)=7*25^n+2*6^(n+1)` is divisible by `19` QED

Answer 2

See below.

Explanation:

We have

`7 cdot 25^n+2 cdot 6^(n+1) =7 cdot (19+6)^n + 12 cdot 6^n`

but

`(19+6)^n = 19^n+n19^(n-1)6+ cdots+n19cdot6^(n-1)+6^n` or

`7 cdot 25^n+2 cdot 6^(n+1) = 7cdot (19k+6^n)+12cdot 6^n=`

`=7 cdot 19 k+19 cdot 6^n equiv 0 mod 19`