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

Answer 1

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,
$$\begin{gathered} \Rightarrow 1+3{}^n{C}_1+3^2{}^n{C}_2+..............+3^{n-1}{}^n{C}_{n-1}+3^n{}^n{C}_n-3n-1 \\ \Rightarrow 1+3n+3^2{}^n{C}_2+..................+3^{n-1}{}^n{C}_{n-1}+3^n{}^n{C}_n-3n-1 \end{gathered}$$
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
$$\begin{gathered} \Rightarrow 3^2\left({}^n{C}_2+................+3^{n-3}{}^n{C}_{n-1}+3^{n-2}{}^n{C}_n\right) \\ \Rightarrow 9\left({}^n{C}_2+................+3^{n-3}{}^n{C}_{n-1}+3^{n-2}{}^n{C}_n\right) \end{gathered}$$
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.