If $$X = \{4^n - 3n - 1 : n \in N\}$$ and $$Y = \{9(n-1) : n \in N\}$$, where $$N$$ is the set of natural numbers, then $$X \cup Y$$ is equal to:

We have two sets of natural-number valued expressions

$$X=\{\,4^{\,n}-3n-1 : n\in N\,\} \quad\text{and}\quad Y=\{\,9(n-1) : n\in N\,\},$$ where $$N=\{1,2,3,\dots\}.$$

Our task is to identify the union $$X\cup Y.$$ To do this we shall show that every element of $$X$$ is already present in $$Y.$$ If that is true then $$X\subseteq Y,$$ and consequently $$X\cup Y=Y.$$

For any natural number $$n,$$ consider the expression

$$4^{\,n}-3n-1.$$

We want to see whether this number is a multiple of $$9.$$ A convenient way to check divisibility by $$9$$ is to look at the remainder (the “modulus”) of the number when divided by $$9.$$ Hence we examine $$4^{\,n}\pmod 9.$$

First, note that

$$4\equiv 4\pmod 9,$$

$$4^{\,2}=16\equiv 7\pmod 9,$$

$$4^{\,3}=64\equiv 1\pmod 9.$$

The remainders repeat every three powers because $$4^{\,3}=1\pmod 9\;\Longrightarrow\;4^{\,3k}=1\pmod 9,$$ and multiplying by another $$4$$ or $$4^{\,2}$$ gives the other two remainders. So we have the periodic pattern

$$4^{\,n}\equiv \begin{cases} 4,& n\equiv 1\pmod 3,\\[6pt] 7,& n\equiv 2\pmod 3,\\[6pt] 1,& n\equiv 0\pmod 3. \end{cases}$$

Now write $$n=3k,\,3k+1,$$ or $$3k+2,$$ and compute the whole expression modulo $$9$$ in each case.

1. If $$n=3k,$$ then $$4^{\,n}\equiv 1.$$ So

$$4^{\,n}-3n-1 =1-3(3k)-1 =1-9k-1 =-9k \equiv 0\pmod 9.$$

2. If $$n=3k+1,$$ then $$4^{\,n}\equiv 4.$$ So

$$4^{\,n}-3n-1 =4-3(3k+1)-1 =4-9k-3-1 =-9k \equiv 0\pmod 9.$$

3. If $$n=3k+2,$$ then $$4^{\,n}\equiv 7.$$ So

$$4^{\,n}-3n-1 =7-3(3k+2)-1 =7-9k-6-1 =-9k \equiv 0\pmod 9.$$

In every case the remainder is zero, so

$$9\;\big\vert\;\bigl(4^{\,n}-3n-1\bigr).$$

Therefore, for each $$n\in N$$ there exists some non-negative integer $$m$$ such that

$$4^{\,n}-3n-1=9m.$$

But the set $$Y=\{9(n-1):n\in N\}$$ contains every non-negative multiple of $$9$$ (namely $$0,9,18,27,\dots$$). Hence the number $$9m$$ we obtained is automatically an element of $$Y.$$ So we have proved

$$X\subseteq Y.$$

Because the union of a set with its superset is the superset itself, we now have

$$X\cup Y=Y.$$

Thus the union equals $$Y,$$ which corresponds to Option B.

Hence, the correct answer is Option B.