Binomial Theorem And Its Simple Applications Question 77
Question: For all $ n\in N,,41^{n}-14^{n} $ is a multiple of
Options:
A) 26
B) 27
C) 25
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Let P(n) be the statement given by $ P(n):41^{n}-14^{n} $ is a multiple of 27
For $ n=1, $
i.e., $ P(1)=41^{1}-14^{1}=27=1\times 27, $
which is a multiple of 27.
$ \therefore P(1) $ is true. Let $ P(k) $ be true, i.e, $ 41^{k}-14^{k}=27\lambda $ (i) For $ n=k+1, $
$ {41^{k+1}}-{14^{k+1}}=41^{k}41-14^{k}14 $
$ =(27\lambda +14^{k})41-14^{k}14 $ [using (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(41\lambda +14^{k}), $
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.
BETA
