Question

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

• A. X
• B. Y
• C. N
• D. None of these

Answer 1

The correct option is B

Y

Since
$4^n-3n-1=(3+1{)}^n-3n-1=3^n+{}^n{C}_1{3}^{n-1}+{}^n{C}_2{3}^{n-2}+.....+{}^n{C}_{n-1}3+{}^n{C}_n-3n-1={}^n{C}_2{3}^2+{}^n{C}_3{.3}^3+...+{}^n{C}_n{3}^n,({}^n{C}_0={}^n{C}_n,{}^n{C}_{n-1}={}^n{C}_1.....soon.)=9[{}^n{C}_2+{}^n{C}_3\left(3\right)+......+{}^n{C}_n{3}^{n-2}]$

$\therefore 4^n-3n-1$ is a multiple of 9 for $n\ge 2$.

For $n=1$,
$4^n-3n-1=4-3-1=0$
For $n=2$,
$4^n-3n-1=16-6-1=9$

$\therefore 4^n-3n-1$ is a multiple of 9 for all $n\in N$
$\therefore X$ contains elements, which are multiples of 9, and clearly $Y$ contains all multiples of 9.

$\therefore X\subset Yi.e.,X\cup Y=Y$