Question

Assertion (A) $P(n):41^n-14^n$ is a multiple of 27 , where $n\in N$.
Reason (R) By principle of mathematical induction, for a given statement $P(n),n\in N$, if $P(1)$ is true and $\bar{P}(k+1)$ is true whenever $P(k)$ is true, $k\in N$, then for all $n\in N,P(n)$ is truc.

• A. Both A and Rare correct; R is the correct explanation of A
• B. Both $A$ and $R$ are correct; $R$ is not the correct explanation of $A$
• C. A is correct: $R$ is incorrect
• D. R is correct; $A$ is incorrect

Answer 1

Answer: (A)

Let the statement $P(n)$ be defined as
$P(n):41^n-14^n$ is a multiple of 27 .
Step I For $n=1$,
$P(1):41^1-14^1=27=1\times 27$
which is a multiple of 27 that is true.
Step II Let it is true for $n=k$,
i.e., $41^k-14^k=27\lambda ....$(i)
Step III For $n=k+1$,
$41^{k+1}-14^{k+1}=41^{k}41-14^{k}14$
$=\left(27\lambda +14^k\right)41-14^{k}14\text{ [using Eq. (i)] }$
$=27\lambda \times 41+14^k\times 41-14^k\times 14$
$=27\lambda \times 41+14^k(41-14)$
$=27\lambda \times 41+14^k\times 27$
$=27\left(41\lambda +14^k\right)$
which is a multiple of 27 .
Therefore, $P(k+1)$ is true when $P(k)$ is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers $n$.
Thus, we see that
Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.