Question

Prove that `4^(2n)-1` is divisible by `5 AA n in NN`?

Answer 1

In the explanation...

Explanation:

To prove this statement by induction, we just have to follow these two steps:
(1) Prove that it holds for `n=1`
(2) Prova that, if it holds for `n-1`, then it should be true for `n`

The first part is as easy as substituting `n=1` on `4^(2n) -1`, which gives us `4^2 - 1 = 16-1 = 15`, and `15` is indeed a multiple of `5`

The second part is a bit more tricky, and we will need to use some simple algebraic manipulation. It goes like this:
Suppose that `4^(2(n-1)) -1` is a multiple of `5`, therefore, there is a natural `k` such that `4^(2(n-1)) -1 = 5k`
Which implies
`4^(2(n-1)) = 5k+1`

`4^(2n-2) = 5k+1`

`4^(2n)*4^(-2)= 5k+1`

`4^(2n)=4^2 *(5k+1)`

`4^(2n)=16 *(5k+1)`

`4^(2n)=80k+16`

`4^(2n)-1= 80k+15`

`4^(2n)-1= 5*(16k+3)`//

And by this, we show that `4^(2n)` is also a multiple of `5` and that the second statement is also correct.
So, by the principle of induction, we have proved that the statement "`4^(2n)-1` is a multiple of `5`" is true for every natural value of `n`.

Hope it helps.

Answer 2

Induction Proof - Hypothesis

We seek to prove that the the expression:

`4^(2n)-1` is divisible by `5 AA n in NN` ..... [A]

So let us test this assertion using Mathematical Induction:

Induction Proof - Base case:

We will show that the given result, [A], holds for `n=1`

When `n=1` the given expression gives:

`4^(2) - 1 = 15`

Which is divisible by `5`, 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:

`4^(2m)-1 = 5k` ..... [B]

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

`4^(2(m+1))-1 -= 4^(2m+2)-1`
`" " = 4^(2m)4^2-1`
`" " = (5k+1)4^2-1` using [B]
`" " = 16(5k+1) - 1`
`" " = 16.5k+16 - 1`
`" " = 16.5k+15`
`" " = 5(16k+3)`

Which is also divisible by `5`

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:

`4^(2n)-1` is divisible by `5` QED