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 $A$.