Question

If $X = \\{ {4^n} - 3n - 1:n \in N\\}$and $Y = \\{ 9(n - 1):n \in N\\}$, then $X \cup Y$ is equal to
$1)X$
$2)Y$
$3)N$
$4)$None of these

Answer 1

First, since from the given Y, there is no possibility of converting the term in any form and we are able to convert the set X.
Convert the given terms in the set of elements of X into the terms of the set of elements of Y.
While converting the set of elements using the Binomial theorem for expanding the terms.
So, use this concept to solve the given question.
Here, N is given as the set of all natural numbers (general assumption).
Formula used: ${(1 + x)^n} = 1 + nx + \dfrac{{n(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + ....$or ${(1 + x)^n} = n{c_0} + {x^n}{c_1} + {x^{2n}}{c_2}...$is the formula for binomial expansion.
The combination formula is $n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}$.

Complete step by step answer:
From the given that the set $Y = \\{ 9(n - 1):n \in N\\}$ and the set $X = \\{ {4^n} - 3n - 1:n \in N\\}$
We need to find the union of these sets, so we will convert the sets into one form, and then it is easy to find the union of the sets.
Let us start converting the set $X = \\{ {4^n} - 3n - 1:n \in N\\}$ where the element $n \in N$.
Set X contains the elements of the form $\Rightarrow {4^n} - 3n - 1$
Rewrite the expression as ${4^n} - 3n - 1 \Rightarrow {(1 + 3)^n} - 3n - 1$
Opening up the bracket terms by using the formula
${(1 + x)^n} = 1 + nx + \dfrac{{n(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + ....$or ${(1 + x)^n} = n{c_0} + {x^n}{c_1} + {x^{2n}}{c_2}...$
Apply the formula is ${(1 + 3)^n}$ then we get,
$\Rightarrow {(1 + 3)^n} = 1 + {3^n}{c_1} + {3^{2n}}{c_2} + ....$
By the use of the combination method we get,
$\Rightarrow {(1 + 3)^n} = 1 + 3n + {3^{2n}}{c_2} + ....$
$\Rightarrow {(1 + 3)^n} - 3n - 1 = (1 + 3n + {3^{2n}}{c_2} + ....) - 3n - 1$
Canceling the common terms, we get
$\Rightarrow {(1 + 3)^n} - 3n - 1 = {3^{2n}}{c_2} + {3^{3n}}{c_3} + ....$
Taking ${3^2}$in common, we get
$\Rightarrow {(1 + 3)^n} - 3n - 1 = {3^2}(^{n}{c_2} + 3.^{n}{c_3} + .....)$
Thus, the set X is converted as
$\Rightarrow {(1 + 3)^n} - 3n - 1 = 9(^{n}{c_2} + 3.^{n}{c_3} + .....)$
Set X has the natural numbers which are multiplied with the $9$(not all), and set Y has the multiples of $9$.
Therefore, $X \subset Y$(the set of X is contained in Y)
So, $X \cup Y$ is equals to the set of all elements in Y (since X is smaller and Y is greater in the set)
Hence, the option $2)Y$ is correct.
Since some of the terms are not in the form of natural numbers hence $3)N$ is incorrect and also $1)X$ is contained in the Y, taking union we get the overall values only, and thus it is incorrect.

So, the correct answer is “Option 2”.

Note: In the given problem the set $X \cup Y$ has the elements of both elements of the sets X and the set Y.
But the elements of the set Y contain the elements of the set X, that is $X \subset Y$.
Since $\subset$is the representation of one set is contained in another set and also called a subset $a \subset b$.
$\supset$is the representation of the reverse process of a subset and denoted as a proper subset which is $a \supset b$.