Question

If $X=\{4^n-3n-1:n\in N\}$ and $Y=\left\{9\right(n-1):n\in N\},$ then show that $X\cap Y=X.$

Answer 1

We have, $X=\{4^n-3n-1:n\in N\}$

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

= $\{1{+}^n{C}_{1}3{+}^n{C}_2{3}^2{+}^n{C}_3{3}^3+...{+}^n{C}_n{3}^n\}-3n-1$

$[\because (1+x{)}^n=1{+}^n{C}_{1}x{+}^n{C}_2{s}^2+...{+}^n{C}_n{x}^n]$

= $1+3n{+}^n{C}_2{3}^2{+}^n{C}_3{3}^3+...{+}^n{C}_n{3}^n-3n-1\left[1\right]$

$[\because {}^n{C}_1=n]$

= ${}^n{C}_2{3}^2{+}^n{C}_3{3}^3+...{+}^n{C}_n{3}^n$

= $3^2[{}^n{C}_2{+}^n{C}_3.3{+}^n{C}_4{3}^2+...{+}^n{C}_n{3}^{n-2}]$

= $9[{}^n{C}_2{+}^n{C}_3.3{+}^n{C}_4{.3}^2+...{+}^n{C}_n{3}^{n-2}]$

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

For $n=1,4^n-3n-1=4-3-1=0$

$\therefore 4^n-3n-1$ is a multiple of $9,\forall n\in N.$

$\therefore$ X contains elements, which are multiple of 9 and clearly Y contains all multiple of 9.

$\therefore x\subseteq Y,i.e.X\cap Y=X$ Hence proved.