Discussion Forum

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

Answer:

Explanation:

We have,

 $X= \left\{  4^{n}-3n-1:n\epsilon N\right\}$

     X= { 0,9,54,243.....}  [put n= 1,2,3 ......]

 $Y= \left\{  9(n-1):n\epsilon N\right\}$

           Y= {0,9,18,27....}

              [Put    n= 1,2,3....]

  It is clear that    $X\subset Y$

$\therefore$      $X\cup Y=Y$