If X = { {4^n} - 3n - 1:n in N} and Y = { 9( | vedclass

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Question

(b) 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} -

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$Y$

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(b) 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$

(${ = ^n}{C_0}={ ^n}{C_n},{^n}{C_1}$ = ${^n}{C_{n - 1}}$ etc.)

$ = 9{[^n}{C_2}{ + ^n}{C_3}(3) + .....{ + ^n}{C_n}{3^{n - 1}}]$

${4^n} - 3n - 1$ is a multiple of $9$ for $n \ge 2$.

For $n = 1,$ ${4^n} - 3n - 1$ = $4 - 3 - 1 = 0$,

For $n = 2,$ ${4^n} - 3n - 1$= $16 - 6 - 1 = 9$

${4^n} - 3n - 1$ is a multiple of $9$ for all $n \in N$

$X$ contains elements, which are multiples of $9$, and clearly $Y$ contains all multiples of $9$.

$X \subseteq Y$ i.e., $X \cup Y = Y$.

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

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