Question

Prove that \[{41^n} - {14^n}\] is a multiple of 27.

Answer 1

Hint:
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.