Question

if $A=\left\{4^n-3n-1:nbelongstoN\right\}$ and $B=\left\{9(n-1):nbelongstoN\right\}$,then

• A. $B\subset A$
• B. $A\cup B=N$
• C. $A\subset B$
• D. None of these

Answer 1

The correct option is D

None of these

The explanation for the correct option:

Step1. The given data:

Given sets

$A=\left\{4^n-3n-1:nbelongstoN\right\}$

and $B=\left\{9(n-1):nbelongstoN\right\}$

Set A can be expressed as

$\begin{array}{rcl}A& =& {(3+1)}^n-3n-1\\ & =& \left({}^{n}C_0{3}^0+{}^{n}C_1{3}^1+{}^{n}C_2{3}^2+\dots .+{}^{n}C_{n-1}{3}^{n-1}+{}^{n}C_n{3}^n\right)-3n-1\\ & =& \left(1+3n+{}^{n}C_2{3}^2+\dots ...+{}^{n}C_n{3}^n\right)-3n-1\\ & =& 9({}^{n}C_2+{}^{n}C_3(3)+\dots ....)\end{array}$

So, $4^n–3n-1$,is a multiple of $9$for $n\ge 2$

Now, Put $n=1,$ in $4^n–3n-1$, we get

$\begin{array}{rcl}{4}^n–3n-1& =& 4-3-1\\ & =& 0\end{array}$

Now, For $n=2$, in $4^n–3n-1$, we get

$\begin{array}{rcl}{4}^n–3n-1& =& 16-6-1\\ & =& 9\end{array}$

Hence, $4^n–3n–1$is a multiple of $9$ for all $n\in N$

Step2. Find the relation on the given sets:

And the set $B=\left\{9(n-1):nbelongstoN\right\}$

Now, Put $n=1,$ in $9(n-1)$, we get

$\begin{array}{rcl}9(n-1)& =& 9(1-1)\\ & =& 0\end{array}$

Now, Put $n=2$ in $9(n-1)$, we get

$\begin{array}{rcl}9(n-1)& =& 9(2-1)\\ & =& 9\end{array}$

Hence, $4^n–3n–1$is a multiple of $9$ for all $n\in N$

$\mathbf{\therefore }\mathbf{}\mathit{A}\mathbf{}\mathbf{\subseteq }\mathbf{}\mathit{B}$ and $\mathit{B}\mathbf{\subseteq }\mathbf{}\mathit{A}$

$\Rightarrow \mathit{A}\mathbf{=}\mathit{B}$

Hence, Option ‘D’ is Correct.