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

JEE Main · 2014 · 06 Apr 2014 · Offline · Mathematical Induction

X = {4^{n} - 3 n - 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

Answer: (D)

Expand with the binomial theorem:

4^{n} = (1 + 3)^{n} = 1 + 3 n + (2 n) 3^{2} + (3 n) 3^{3} + \dots

, so

4^{n} - 3 n - 1 = (2 n) 3^{2} + (3 n) 3^{3} + \dots = 9 ((2 n) + 3 (3 n) + \dots)

. Every member of

X

is thus a multiple of

9

, while

Y = {9 (n - 1)} = {0, 9, 18, \dots}

is the set of all multiples of

9

. Since

X \subseteq Y

, the union

X \cup Y

is just

Y

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option