Binomial-Theorem-And-Its-Simple-Applications Question 233
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
Correct Answer: B $ \therefore P(1) $ is true.
Let $ P(k) $ be true, i.e, $ 41^{k}-14^{k}=27\lambda $ (i)
For $ n=k+1, $Show Answer
Answer:
Solution:
i.e., $ P(1)=41^{1}-14^{1}=27=1\times 27, $
which is a multiple of 27.
$ {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
