Using the principle of mathematical induction prove that `41^n-14^n` is a multiple of `27` .

Explanation

`P(n) : 41^(n)-14^(n) = 27x, x in N`
For `n = 1 , 41^(1) - 14^(1) = 27`
So, `P(1)` is true,
Let `P(k)` be true.
i.e, `41^(k) - 14^(k) = 27m, m in N"….."(1)`,
Now, we have to prove that `P(k+1)` is true.
i.e, `41^(k+1)-14^(k+1)=27p, p in N`
`41^(k+1)-14^(k+1)`
`= 41 xx 41^(k) - 14 xx 14^(k)`
` = 41 xx (41^(k) - 14^(k) + 14^(k) xx 14)`
`= 41(41^(k) - 14^(k)) + 41 xx 14^(k) - 14^(k) xx 14`
` = 41 xx 27 m + 27 xx 14^(k)` [Using `(1)`]
`= 27 xx (41m + 14^(k))`
`= 27 xx q [ q = (41m + 14^(k))]`
thus, `P(k+1)` is true whenever `P(k)` is true.
Hence, by the principle of mathematical induction, statement `P(n)` is true for all natural numbers.