prove that `4^(n)-1` is divisible by 3, for each natural number n.

Explanation

Let `P(n):4^(n)-1` is divisible by 3 for each natural number n. Step I Now, we observe that P(1) is true.
`P(1)=4^(1)-1=3`
It is clear that 3 is divisible by 3.
Hence, P(1) is true.
Step II Assume that, P(n) is true for n=k
`P(k):4^(k)-1` is divisible by 3
`x4^(k)-1=3q`
Step III Now, to prove that P(k+1) is true. `P(k+1):4^(k+1)-1`
`=4^(k).4-1`
`=4^(k).3+4^(k)-1`
`=3*4^(k)+3q[because4^(k)-1=3q]`
`=3(4^(k)+q)`
Thus, P(k+1) is true whenever P(k) is true.
Hence, by the principle of mathematical P(n) is true for all natural number n.