Binomial Theorem And Its Simple Applications Question 71
Question: For every positive integer n, $ 7^{n}-3^{n} $ is divisible by
Options:
A) 7
B) 3
C) 4
D) 5
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Let $ P(n):7^{n}-3^{n} $ is divisible by 4. For $ n=1 $ , $ P(1):7^{1}-3^{1}=4, $ which is divisible by 4. Thus, P (n) is true for n = 1. Let P (k) be true for some natural number k, i.e. $ P(k):7^{k}-3^{k} $ is divisible by 4. We can write $ 7^{k}-3^{k}=4d, $ where $ d\in N $ - (i) Now, we wish to prove that $ P(k+1) $ is true whenever P(k) is true, i.e., $ {7^{k+1}}-{3^{k+1}} $ is divisible by 4. Now, $ {7^{(k+1)}}-{3^{(k+1)}} $
$ ={7^{(k+1)}}-{{7.3}^{k}}+{{7.3}^{k}}-{3^{(k+1)}} $
$ =7(7^{k}-3^{k})+(7-3)3^{k}=7(4d)+{{4.3}^{k}} $ [Using (i)] $ =4(7d+3^{k}), $ Which is divisible by 4. Thus, $ P(k+1) $ is true whenever $ P(k) $ is true. Therefore, by the principle of mathematical induction the statement is true for every positive integer n.
BETA
