Question

Consider two sets $X$and $Y$such that,
$X = \\{ {8^n} - 7n - 1:n \in N\\}$and $Y = \\{ 49(n - 1):n \in N\\} ,$ then choose the correct option:
(A) $X = Y$ (B) $X \subset Y$ (C) $Y \subset X$ (D) None of these

Answer 1

Hint: Use ${8^n} = {(1 + 7)^n}$ and then apply binomial expansion.
The elements in $X$ are in the form of ${8^n} - 7n - 1$. So, this can be written as:
${8^n} - 7n - 1 = {(1 + 7)^n} - (1 + 7n)$
Using binomial expansion for${(1 + 7)^n}$:

$$\Rightarrow {8^n} - 7n - 1{ = ^n}{C_0}{ + ^n}{C_1} \times 7{ + ^n}{C_2} \times {7^2}{ + ^n}{C_3} \times {7^3} + .....{ + ^n}{C_n} \times {7^n} - (1 + 7n) \\\ \Rightarrow {8^n} - 7n - 1 = 1 + 7n{ + ^n}{C_2} \times {7^2}{ + ^n}{C_3} \times {7^3} + .....{ + ^n}{C_n} \times {7^n} - (1 + 7n) \\\ \Rightarrow {8^n} - 7n - 1{ = ^n}{C_2} \times {7^2}{ + ^n}{C_3} \times {7^3} + .....{ + ^n}{C_n} \times {7^n} \\\ \Rightarrow {8^n} - 7n - 1 = {7^2}{(^n}{C_2}{ + ^n}{C_3} \times 7 + .....{ + ^n}{C_n} \times {7^{n - 2}}), \\\ \Rightarrow {8^n} - 7n - 1 = 49{(^n}{C_2}{ + ^n}{C_3} \times 7 + .....{ + ^n}{C_n} \times {7^{n - 2}}) \\\$$

Clearly, ${8^n} - 7n - 1$ will be a multiple of $49$ for different values of $n.$
$\therefore$Set $X$ contains some multiple of $49$
Elements of $Y$on the other hand are represented by $49(n - 1)$for all natural numbers.
$\therefore$Set $Y$contains all multiples of $49$ including $0$
Clearly set $X$ is a proper subset of set $Y$
Therefore, $X \subset Y$is the correct relation between the two sets and option (B) is correct.
Note: If two sets $A$ and $B$ are such that set $B$ contains all elements of set $A$ along with at least one extra element of itself, then $A$ is said to be the proper subset of $B$ and is denoted by $A \subset B.$