Question

{(41)^n - (14)^n) is divisible by 27.

using principal of Mathematical induction prove this​

Answer 1

We have to prove,

$P(n)\ :\ \ 27\ \mid\ 41^n-14^n,\ \ \forall n\in\mathbb{N}$

First, let's check whether P(1) is true or not.

41^1 - 14^1  =  41 - 14  =  27

Here,  27 | 27.  So P(1) is true.

Hence we can assume that P(k) is true.

$\textsf{Assume that\ \ P(k)\ :\ 27\ \mid\ 41^k-14^k \ \ holds true.}$

So we have to write it as a multiple of 27.

$\textsf{Let}\ \ \ 41^k-14^k=27m,\ \ m\in\mathbb{N}$

This implies,

$\[\]\\ \begin{equation}41^k=14^k+27m\end{equation}$

Now let's consider  P(k + 1).

$\begin{aligned}&P(k+1)\ :&&41^{k+1}-14^{k+1}\\ \\ &\Longrightarrow&&41^k\cdot 41-14^k\cdot 14\\ \\ &\Longrightarrow&&(14^k+27m)41-14^k\cdot 14\ \ \ \ \ \ \ \ \ \ [\ \text{From\ \ (1)}\ ]\\ \\ &\Longrightarrow&&14^k\cdot 41+27m\cdot 41-14^k\cdot 14\\ \\ &\Longrightarrow&&14^k(41-14)+27m\cdot 41\\ \\ &\Longrightarrow&&14^k\cdot 27+27m\cdot 41\\ \\ &\Longrightarrow&&27(14^k+41m)\end{aligned}$

This implies that P(k + 1) also holds true because the last step derives it as a multiple of 27.

Hence Proved!