Prove that 7n−17n−1 is a multiple of 6 for all n∈N.

Solution:

Let P(n) be $7^{n}-1$ is a multiple of 6 for all n∈N.

For n=1

P(1): $7^{1}-1=7-1=6$ is a multiple of 6.

So, P(1) is true.

Now,we assume that P(k) is true.

P(k): $7^{k}-1$ is a multiple of 6

$\Rightarrow 7^k-1=6m$ [for some constant m] ...(i)

Now, we show that P(k+1) is true.

P(k+1): $7^{k+1}-1$ is a multiple of 6.

Take $7^{k+1}-1$

$=7^k7-1 \\=(6m+1)7-1\ \ \ [from (i)] \\=42m+7-1 \\=42m+6 \\=6(7m+1)$

which is clearly a multiple of 6.

Hence, by the principle of mathematical induction, given statement P(n) is true for all n ∈N