Question

If \[X = \left \{ {{8^n} - 7n - 1,n \in N} \right \} \] and \[Y = 49(n - 1) \] , \[n \in N \] , then(given \[n > 1 \] )
A. \[X \subset Y \]
B. \[Y \subset X \]
C. \[X = Y \]
D. \[X \varsubsetneq Y \]

Answer 1

Hint: Since options are given in this question. We can solve this easily by putting the values of n in above X and Y. We can also solve this question using the binomial expansion. Remember the definition of subset. That is a set A is a subset of B if all elements of the set will be in the set B.

Complete step-by-step answer:
Given, \[X = \left \{ {{8^n} - 7n - 1,n \in N} \right \} \] and \[Y = \left \{ {49(n - 1)|n \in N} \right \} \]
Method (1):
Putting \[n = 1,2,3,4,5,.., \] we have, in X.
 \[\Rightarrow X = \{ 0,49,490,4067,32732,..... \} \]
Randomly we put, \[n = 1,2,3,.....,11,....,84,.....,669,... \] in Y.
 \[\Rightarrow Y = \{ 0,49,98,...,490,....,4067,...,32732,... \} \]
By the definition of subset we can see that \[X \subset Y \] .
Method (2):
Consider, \[{8^n} - 7n - 1 \]
8 can be written as 1+7,
 \[ = {(1 + 7)^n} - 7n - 1 \]
We know binomial expansion
 \[{(1 + x)^n} = 1 + \dfrac{n}{{1!}}x + \dfrac{{n(n - 1)}}{{3!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + - - - - + {x^n} \]
Using this above becomes,
 \[ = 1 + 7n + \dfrac{{n(n - 1)}}{{2!}}{7^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{7^3} + - - - - - + {7^n} - 7n - 1 \]
Cancelling the terms, we get
 \[ = \dfrac{{n(n - 1)}}{{2!}}{7^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{7^3} + - - - - - \]
Taking 49 as common, we get:
 \[ = {7^2} \left( { \dfrac{{n(n - 1)}}{{2!}} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}7 + - - - - - } \right) \]
Obviously the terms in the bracket will be positive for n=3,4,….
Hence, we get,
 \[ = 49({ \text{some positive number)}} \]
When \[{ \text{n = 1}} \] , then, \[{8^n} - 7n - 1 = {8^1} - 7.1 - 1 = 0 \] , meaning that 0 is a multiple 0f 49.
Hence \[49|({8^n} - 7n - 1) \] , \[ \forall n \in N \] .
This implies \[X \] contains only the multiples of (Not all) 49.
But it is evident that \[Y = 49(n - 1) \] contains all of the multiples of 49.
By the definition of sunsets we conclude that \[X \subset Y \] .
In both cases we have the same answer. \[X \subset Y \]
So, the correct answer is “Option A”.

Note: I recommend you to follow method (2). Method (1) seems easy but needs a lot of multiplication and addition. Which consumes a lot of time. We can also solve this by using mathematical induction, but it is quite difficult to solve compared to the above two methods. Remember the binomial expansion.