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 $Y$

Set $X$ contains elements of the form

$4^n-3n-1=(1+3{)}^n-3n-1$

$=3^n{+}^n{C}_{(}n-1)3^{n-1}{\cdots }^n{C}_2{3}^2$

$=9\left(3^{n-2}{+}^n{C}_{(n-1)3^{n-1}}{\cdots }^n{C}_2\right)$

Set $X$ has natural numbers which are multiples of $9$

Set $Y=9(n-1)$

Which implies Set $Y$ has all multiples of $9$

$X\cup Y=Y$