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1.Set Theory
hard
If $X = \{ {4^n} - 3n - 1:n \in N\} $ and $Y = \{ 9(n - 1):n \in N\} ,$ then $X \cup Y$ is equal to
A
$X$
B
$Y$
C
$N$
D
None of these
Solution
(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., $X \cup Y = Y$.
Standard 11
Mathematics
