Consider the given statement is
`P(n):7^(n)-2^(n)` is
Step I We observe that P(1) is true.
`P(1)=7^(1)-2^(1)=5`, which is disivible by 5.
Step II Now, assume that P(n) is true for n=k.
`P(k)=7^(k)-2^(k)=5q`
Step III Now, to prove P(k+1) is true,
`P(k+1):7^(k+1)-2^(k+1)`.
`=7^(k)*2^(k)*2`
`=7^(k)*(5+2)-2^(k)*2`
`=7^(k)*5+2*7^(k)-2^(k)*2`
`5*7^(k)+2(7^(k)-2^(k))`
`=5*7^(k)+2(5q)`
`=5(7^(k)+2q)`, which is divisible by 5. [from step II]
So, P(k+1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true for any natural number n.
