Binomial Theorem And Its Simple Applications Question 66
Question: By the principle of induction $ \forall n\in N,3^{2n} $ when divided by 8, leaves remainder
Options:
A) 2
B) 3
C) 7
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Let P (n) be the statement given by
$ P(n):3^{2n} $ when divided by 8, the remainder is 1. or $ P(n):3^{2n}=8\lambda +1 $ for some $ \lambda \in N $
For $ n=1,,P(1):3^{2}=(8\times 1)+1=8\lambda +1, $ where $ \lambda =1 $
$ \therefore ,P(1) $ is true. Let P (k) be true. Then $ 3^{2k}=8\lambda +1 $ for some $ \lambda \in N $ - (i) We shall now show that $ P(k+1) $ is true, for which we have to show that $ {3^{2(k+1)}} $ when divided by 8, the remainder is 1. Now $ {3^{2(k+1)}}=3^{2k}{{.3}^{2}} $
$ =(8\lambda +1)\times 9 $ [Using (i)] $ =72\lambda +9=72\lambda +8+1=8(9\lambda +1)+1 $
$ =8\mu +1, $ where $ \mu =9\lambda +1\in N $
$ \Rightarrow P(k+1) $ is true. Thus, $ P(k+1) $ is true, whenever P(k) is true.
Hence, by the principle of mathematical induction $ P(n) $ is true for all $ n\in N $ .
BETA
