Question

If we are given two sets as $A=\{4^n-3n-1:n\in N\}\text{ and }B=\{9(n-1):n\in N\}$ then?
$$\begin{array}{rl}& A.B\subset A\\ & B.A\cup B=N\\ & C.A\subset B\\ & D.\text{None of these}\end{array}$$

Answer 1

Hint: In this question, we are given a set builder form of two sets A and B. We need to check the options and answer if B is a subset of A or A is a subset of B or union of A and B form a set of natural numbers or none of the options correct. For this, we will simplify the set A and B and then check options. For simplification we will use formula of binomial expansion given by ${(1+x)}^n=1+nx+{}^n{C}_2{x}^2+{}^n{C}_3{x}^3+\dots \dots +{}^n{C}_n{x}^n$ where ${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$.

Complete step-by-step solution
Here we are given two sets A and B. Set A is given by $A=\{4^n-3n-1:n\in N\}$ and set B is given by $B=\{9(n-1):n\in N\}$.
For finding our answer, let us simplify the given sets.
Simplifying set A first:
Taking the expression $4^n-3n-1$ and simplifying it, we get ${(1+3)}^n-3n-1$.
Now let us apply binomial expansion on ${(1+3)}^n$. As we know, binomial expansion on ${(1+x)}^n$ is given by ${(1+x)}^n=1+nx+{}^n{C}_2{x}^2+{}^n{C}_3{x}^3+\dots \dots +{}^n{C}_n{x}^n$.
So putting x=3 we get:
${(1+3)}^n=1+3n+{}^n{C}_2{\left(3\right)}^2+{}^n{C}_3{\left(3\right)}^3+\dots \dots +{}^n{C}_n{\left(3\right)}^n$.
Now, ${(1+3)}^n-3n-1$ becomes equal to $1+3n+{}^n{C}_2{\left(3\right)}^2+{}^n{C}_3{\left(3\right)}^3+\dots \dots +{}^n{C}_n{\left(3\right)}^n-3n-1$.
Simplifying we get: ${}^n{C}_2{\left(3\right)}^2+{}^n{C}_3{\left(3\right)}^3+\dots \dots +{}^n{C}_n{\left(3\right)}^n$.
Now taking ${\left(3\right)}^2$ common from all terms, we get:
$\begin{array}{rl}& 3^2({}^n{C}_2+{}^n{C}_3\left(3\right)+\dots \dots +{}^n{C}_n{\left(3\right)}^{n-2})\\ & \Rightarrow 9({}^n{C}_2+3{}^n{C}_3+\dots \dots +{\left(3\right)}^{n-2}{}^n{C}_n)\end{array}$
Now set A becomes of the form:
$A=\{9({}^n{C}_2+3{}^n{C}_3+\dots \dots +{\left(3\right)}^{n-2}{}^n{C}_n):n\in N\}$
As we can see, all elements of set A will be some multiple of 9. Let us find some elements of A,
Putting n = 1, we get 0.
Putting n = 2, we get 9.
Putting n = 3, we get 54.
So set $A=\{0,9,54,\dots \dots \}$ (All numbers are some multiple of 9).
Now let us look into set B.
$B=\{9(n-1):n\in N\}$.
Set B is already simplified and putting values of n as 1, 2, 3 ..... will give us all multiple of 9.
Hence, $B=\{0,9,18,27,\dots \dots \}$.
As we can see, set A has some multiple of 9, and set B is a set consisting of all multiple of 9. Hence, set A is a subset of set B.
Hence, $A\subset B$.
Therefore, option C is the correct answer.

Note: Students should take care while applying the binomial expansion. General form of binomial expansion is given by ${(a+b)}^n=a^n+n{a}^{n-1}b+{\displaystyle \frac{n(n-1)}{2!}}{a}^{n-2}{b}^2+\dots \dots +b^n$.
Students should know the notations $\subset ,\supset$ for solving these sums. $A\subset B$ means A is a subset of B whereas $A\subset B$ means A is a subset of B (or B is a subset of A). When we write sets in form of exact elements, then it is called roaster form and when we write the general way of giving elements of the set, then it is called set builder form.