**4n – 1 is divisible by 3, for each natural number .**

###### Answers:

Let P(n): 4^{n }– 1 is divisible by 3 for each natural number n.

Now, P(l): 4^{1} – 1 = 3, which is divisible by 3 Hence, P(l) is true.

Let us assume that P(n) is true for some natural number n = k.

P(k): 4^{k }– 1 is divisible by 3

or 4^{k} – 1 = 3m, m∈ N (i)

Now, we have to prove that P(k + 1) is true.

P(k+ 1): 4^{k+1} – 1

= 4^{k}-4-l

= 4(3m + 1) – 1 [Using (i)]

= 12 m + 3

= 3(4m + 1), which is divisible by 3 Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction P(n) is true for all natural numbers n.