If X=4^n-3n-1:n N and Y=9(n-1):n N, where N is the set of natural numbers, then X Y is equ…

Question

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

Accepted Answer

$$4^n-3n-1=(1+3)^n-3n-1={}^nC_0+{}^nC_1\cdot3+{}^nC_2\cdot3^2+\cdots+{}^nC_n3^n-3n-1$$

$$=9\left[{}^nC_2+{}^nC_3\cdot3+\cdots+{}^nC_n3^{n-2}\right]$$

So $4^n-3n-1$ is always a multiple of 9: $X=\{\text{multiples of }9\}$. Also $Y=\{9(n-1):n\in N\}=\{\text{all multiples of }9\}$. Since $X\subset Y$ (and in fact $X\subseteq Y$), $X\cup Y=Y$.

Solutions · 1