Sets Relations And Functions Question 63
Question: If $ X={8^{n}-7n-1:n\in N} $ and $ Y={49(n-1):n\in N}, $ then
Options:
A) $ X\subseteq Y $
B) $ Y\subseteq X $
C) $ X=Y $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- Since $ 8^{n}-7n-1={{(7+1)}^{n}}-7n-1 $ $ =7^{n}{{+}^{n}}C_1{7^{n-1}}{{+}^{n}}C_2{7^{n-2}}+…..{{+}^{n}}{C_{n-1}}7{{+}^{n}}C_{n}-7n-1 $ $ {{=}^{n}}C_27^{2}{{+}^{n}}C_37^{3}+..{{+}^{n}}C_{n}7^{n} $ , $ {{(}^{n}}C_0{{=}^{n}}C_{n},{{,}^{n}}C_1{{=}^{n}}{C_{n-1}},etc\text{.)} $ $ =49{{[}^{n}}C_2{{+}^{n}}C_3(7)+……{{+}^{n}}C_{n}{7^{n-2}}] $
$ \therefore $ $ 8^{n}-7n-1 $ is a multiple of 49 for $ n\ge 2 $ For $ n=1 $ , $ 8^{n}-7n-1=8-7-1=0 $ ; For $ n=2, $ $ 8^{n}-7n-1=64-14-1=49 $
$ \therefore $ $ 8^{n}-7n-1 $ is a multiple of 49 for all $ n\in N. $
$ \therefore $ X contains elements which are multiples of 49 and clearly Y contains all multiplies of 49.
$ \therefore $ $ X\subseteq Y $ .
BETA
