Question

# If $X = \left\{ {{4^n} - 3n - 1:n \in N} \right\}$and $Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\}$, then $X \cup Y$ is equal to A. $X$B. $Y$C. $N$D. None of the above

Hint: Convert the set of elements of $X$ in terms of the set of elements of $Y$. While converting the set of elements use Binomial Theorem for expanding the terms. So, use this concept to reach the solution of the problem.

Given set $Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\}$ and
Set $X$ contains elements of the form
$\Rightarrow {4^n} - 3n - 1$
Which can be written as
$\Rightarrow {\left( {1 + 3} \right)^n} - 3n - 1$
Opening the terms in the bracket by using the formula ${\left( {1 + x} \right)^n} = {}^n{C_0} + x{}^n{C_1} + {x^2}{}^n{C_2} + ....................... + {x^{n - 1}}{}^n{C_{n - 1}} + {x^n}{}^n{C_n}$ we have,
$\Rightarrow 1 + 3{}^n{C_1} + {3^2}{}^n{C_2} + .............. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n} - 3n - 1 \\ \Rightarrow 1 + 3n + {3^2}{}^n{C_2} + .................. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n} - 3n - 1 \\$
Cancelling the common terms, we get
$\Rightarrow {3^2}{}^n{C_2} + .................. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n}$
Taking ${3^2}$as common, we get
$\Rightarrow {3^2}\left( {{}^n{C_2} + ................ + {3^{n - 3}}{}^n{C_{n - 1}} + {3^{n - 2}}{}^n{C_n}} \right) \\ \Rightarrow 9\left( {{}^n{C_2} + ................ + {3^{n - 3}}{}^n{C_{n - 1}} + {3^{n - 2}}{}^n{C_n}} \right) \\$
Clearly, set $X$ has natural numbers which are multiples of 9 (not all) and the set $Y$has all the multiples of 9.
Therefore, $X \subset Y$. So $X \cup Y$ is equal to the set of elements in $Y$.
Thus, the correct option is B. $Y$

Note: In the given problem the set $X \cup Y$ has the elements of both elements of the sets$X$ and $Y$. But the elements of set $Y$ contain the elements of set $X$ i.e., $X \subset Y$ from the solution.