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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

Answer Verified Verified
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.

Complete step-by-step answer:
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.
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