If X-4n-3 n-1- n - N and Y-9n-1- n - N - where N is the set of natural numbers- then X - Y is equal to-A- YB- NC- Y - XD- X

The correct option is A Y4^n 3n 1 = (1+3)^n 3n 1 =[^nC_0 + ^nC_1 .3+^nC_2. 3^2 + ⋯ +^nC_n 3^n] 3n 1 =9[^nC_2 + ^nC_3.3 + ⋯ +^nC_n.3^n 2] ∴ 4^n 3n 1 is a ...

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Byju's Answer

Standard XII

Mathematics

Subset

If X=4n-3 n-1...

Question

If $X = {4^{n} - 3 n - 1 : n ϵ N} a n d Y = {9 (n - 1) : n ϵ N}$ , where N is the set of natural numbers, then $X \cup Y$ is equal to:

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Y - X

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Solution

The correct option is B Y
$4^{n} - 3 n - 1 = (1 + 3)^{n} - 3 n - 1 = [^{n} C_{0} +^{n} C_{1} .3 +^{n} C_{2} . 3^{2} + \dots +^{n} C_{n} 3^{n}] - 3 n - 1 = 9 [^{n} C_{2} +^{n} C_{3} .3 + \dots +^{n} C_{n} {.3}^{n - 2}]$
$∴ 4^{n} - 3 n - 1$ is a multiple of 9 for all n.
$∴$ X = {x : x is a multiple of 9}
Also, $Y = 9 (n - 1) : n ϵ N =$ {All multiples of 9}
Clearly $X \subseteq Y . ∴ X \cup Y = Y$

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If X={4n−3n−1 : nϵN} and Y={9(n−1):nϵN}, where N is the set of natural numbers, then X∪Y is equal to:

The correct option is B Y4n−3n−1=(1+3)n−3n−1=[nC0+nC1.3+nC2.32+⋯+nCn3n]−3n−1=9[nC2+nC3.3+⋯+nCn.3n−2] ∴ 4n−3n−1 is a multiple of 9 for all n. ∴ X = {x : x is a multiple of 9} Also, Y=9(n−1):nϵN= {All multiples of 9} Clearly X⊆Y. ∴ X∪Y=Y

If $X = {4^{n} - 3 n - 1 : n ϵ N} a n d Y = {9 (n - 1) : n ϵ N}$ , where N is the set of natural numbers, then $X \cup Y$ is equal to: