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

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Vedantu Content Team · Accepted Answer

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.

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

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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

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