Using the principle of mathematical induction , prove that for `n in N`, `41^(n) - 14^(n)` is a multiple of 27.

Using the principle of mathematical induction , prove that for `n in N

1.	Using the principle of mathematical induction , prove that for `n in N`, `41^(n) - 14^(n)` is a multiple of 27.
Answer» `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.