Question

If $X = \\{4^n- 3 n - 1: n \in N \\}$ and $Y = \\{9 (n - 1): n \in N \\}$ ,where $N$ is the set of natural numbers, then $X \cup Y$ is equal to

• A. N
• B. Y-X
• C. X
• D. Y

Answer 1

Answer: (D)

Y

Answer 2

X=\left\\{(1+3)^{n}-3 n-1, n \in N\right\\}
=3^{2}({ }^{n} C_{2}+{ }^{n} C_{3} \cdot 3+\ldots+3^{n-2}), n \in N\\}
$=\\{\text { Divisible by } 9\\}$
$Y=\\{9(n-1), n \in N\\}$
=(\text { All multiples of } 9\\}
So, $X \subseteq Y$
i.e., $X \cup Y=Y$