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Prove the following by using the principle of mathematical induction for all $n \in N$ \[\] $ 41^n-14^n$ is a multiple of 27.

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Let the given statement be $P(n)$, i.e.,
$P(n) : 41^n-14^n$ is a multiple of 27.
It can be observed that $P(n) $ is true for $n=1$ Since $41^1-14^1=27$, which is divisible by 27.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$ 41^k-14^k $ is a multiple of 27.
$ \therefore 41^k-14^k =27m$, where $m \in N$-----------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
$ = 41^k.41-14^k.14$
$ =41(41^k-14^k+14^k)-14^k.14$
$ =41.27m+14^k(41-14)$
$= 41.27m+27.14^k$
$ = 27(41m-14^k)$
$= 27 \times r$, where $r=(41m-14^k)$ is a natural number.
Therefore, $41^{k+1}-14^{k+1}$ is a multiple of 27.
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 i.e., $n$.
answered May 1, 2014 by thanvigandhi_1

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