Question

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

• A. $N$
• B. $Y-X$
• C. $X$
• D. $Y$

Answer 1

The correct option is D $Y$

$X=\{4^n-3n-1:n\in N\}$
$=(1+3{)}^n-3n-1$
$={}^n{C}_0{3}^0+{}^n{C}_1{3}^1+{}^n{C}_2{3}^2+...+{}^n{C}_n{3}^n-3n-1$
$=9[{}^n{C}_2+{}^n{C}_{3}3+{}^n{C}_4{3}^2+...+{}^n{C}_{n-2}{3}^{n-2}]$
$\therefore X$ is multiple of $9$
So, $X=0,9,54,........\left(1\right)$
Now, $Y=\left\{9\right(n-1):n\in N\}$
$\therefore Y$ is also multiple of $9$
So, $Y=0,9,\mathrm{18................}\left(2\right)$
From equation (1) and (2)
$X\cup Y=Y$