If X = 4n- 3 n - 1: n ∈ N and Y = 9 (n - 1): n ∈ N ,where N is the set of natural numbers, then X ∪ Y is equal to

Question

If X = 4n- 3 n - 1: n ∈ N and Y = 9 (n - 1): n ∈ N ,where N is the set of natural numbers, then X ∪ Y is equal to (A) N (B) Y-X (C) X (D) Y

Tardigrade Admin · Accepted Answer

X= (1+3)n-3 n-1, n ∈ N =32( n C2+ n C3 ⋅ 3+ ldots+3n-2), n ∈ N = text Divisible by 9 Y= 9(n-1), n ∈ N =( text All multiples of 9 So, X ⊆ Y i.e., X ∪ Y=Y

Q. 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

N

Y-X

X

Y

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

Q. If X={4n−3n−1:n∈N} and Y={9(n−1):n∈N} ,where N is the set of natural numbers, then X∪Y is equal to

N

Y-X

X

Y

X={(1+3)n−3n−1,n∈N} =32(nC2​+nC3​⋅3+…+3n−2),n∈N} ={ Divisible by 9} Y={9(n−1),n∈N} =( All multiples of 9} So, X⊆Y i.e., X∪Y=Y

Q. 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

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