Question

Prove that ${41^n} - {14^n}$ is a multiple of 27.

Answer 1

Here we will use the concept of the principle of mathematical induction. We will show that the given expression is true for $n = 1$. Then, we will assume that the given expression is true for all real numbers. Finally, we will use it to prove the given statement.

Complete step by step solution:
Here we have to prove ${41^n} - {14^n}$ appears in the multiplication table of 27. We will do this by the principal of mathematical induction.
Let $P\left( n \right) = {41^n} - {14^n}$
Let us first show that $P\left( n \right)$ is true for $n = 1$ i.e., $P\left( 1 \right)$ is a multiple of 27.
Now, $P\left( 1 \right) = {41^1} - {14^1} = 27$.
We know that 27 is a multiple of 27. So, $P\left( 1 \right)$ is true.
Next, we will assume that $P(k)$ is true for all $k \in \mathbb{N}$i.e., we will assume that ${41^k} - {14^k}$is a multiple of 27.
Let ${41^k} - {14^k} = 27m$, where $m \in \mathbb{N}$
Adding ${14^k}$ on both sides, we get
${41^k} = 27m + {14^k}$……….$(1)$
Now, we have to prove that $P(k + 1)$ is true i.e., we have to show that ${41^{k + 1}} - {14^{k + 1}}$ is a multiple of 27.
Let us consider the expression ${41^{k + 1}} - {14^{k + 1}}$.
We can write this as
${41^{k + 1}} - {14^{k + 1}} = 41 \times {41^k} - 14 \times {14^k}$
Substituting equation $(1)$ in above equation, we have
$\Rightarrow {41^{k + 1}} - {14^{k + 1}} = 41 \times (27m + {14^k}) - 14 \times {14^k}$
Multiplying the terms on the RHS, we get
$\Rightarrow {41^{k + 1}} - {14^{k + 1}} = 41 \times 27m + 41 \times {14^k} - 14 \times {14^k}$
Now taking ${14^k}$ as common, we get
$\Rightarrow {41^{k + 1}} - {14^{k + 1}} = 41 \times 27m + (41 - 14) \times {14^k}$
We know that $41 - 14 = 27$. Hence, above equation becomes
$\Rightarrow {41^{k + 1}} - {14^{k + 1}} = 41 \times 27m + 27 \times {14^k}$
Taking 27 common, we get
$\Rightarrow {41^{k + 1}} - {14^{k + 1}} = 27(41m + {14^k})$
Let us take $41m + {14^k} = r$, where $r \in \mathbb{N}$. Therefore, the above equation becomes
$\Rightarrow {41^{k + 1}} - {14^{k + 1}} = 27r$
The observation in the above equation is a multiple of 27, which means that ${41^{k + 1}} - {14^{k + 1}}$ is a multiple of 27. Hence, $P(k + 1)$ is true.

Therefore, by the principle of mathematical induction, $P(n)$ is true for all $n \in \mathbb{N}$and so ${41^n} - {14^n}$ is a multiple of 27.

Note:
Mathematical Induction is a technique of proving a statement, theorem or formula which is thought to be true, for each and every natural number $n$. In the given problem, if $n = 1$, then ${41^n} - {14^n}$ is a multiple of 27. So, to prove it for all $n \in \mathbb{N}$, we adopt the principle of mathematical induction. If $n = 1$ and ${41^n} - {14^n}$ is not a multiple of 27 then we cannot prove it using the principle of mathematical induction.