Question

If \[X=\left\{ {{4}^{n}}-3n-1:n\in N \right\}\]and \[Y=\left\{ 9\left( n-1 \right):n\in N \right\}\], where N is the set of natural numbers, then \[X\cup Y\] is equal to
A. Y
B. Y-X
C. X
D. N

Answer 1

Hint: we will obtain the elements in set X by using binomial expansion of \[{{\left( 1+x \right)}^{n}}\]. This will help you to reach the solution.

Complete step by step answer:
Set X contains elements of the form
\[\begin{align}
  & {{4}^{n}}-3n-1 \\
 & ={{\left( 1+3 \right)}^{n}}-3n-1 \\
 & ={{3}^{n}}+{}^{n\mathop{c}_{n-1}}{{3}^{n-1}}+........+{}^{n\mathop{c}_{2}}{{3}^{2}}+{}^{n\mathop{c}_{1}}{{3}^{1}}+1-3n-1 \\
\end{align}\]
=\[{{3}^{n}}+{}^{n\mathop{c}_{n-1}}{{3}^{n-1}}+........+{}^{n\mathop{c}_{2}}{{3}^{2}}\] \[\]
\[=9\left( {{3}^{n-2}}+{}^{n\mathop{c}_{n-1}}{{3}^{n-3}}+........+{}^{n\mathop{c}_{2}}{{3}^{0}} \right)\]
We will get the above expression by cancelling the terms positive 3n and negative 3n, cancelling positive 1 and negative 1 and from the remaining expression take 9 common from each term and hence we can get the above expression.
So we can conclude that set X has natural numbers which are multiples of 9 because we get set X as multiplication of 9 and some factor.so we can say that set X has multiples of 9 but it doesn’t contain all multiples of 9 as known from the above expansion we derived.
Set Y contains elements of the form
\[Y=\left\{ 9\left( n-1 \right) \right\}\]
From the above expansion it is clear that set Y contains all multiples of 9.
Check one by substituting the value of n by natural numbers and we will get 0,9,18,27, 36, 45,54 . . . . . . and hence we can conclude that set Y contains all the multiples of 9.
Now we want \[X\cup Y\]
If set X is a subset of set Y then \[X\cup Y\]=\[Y\]

So, the correct answer is “Option A”.

Note: if there are two sets present as set A and set B and set B is a subset of set A then the union of set A and set B is set A because all elements in set B are present in set A and some more elements are present in set A so union of those two sets is set A.