If X = { 4 n - 3n - 1 : n ∈ N} and Y = { 9(n-1) : n ∈ N}, then X ∪ Y is equal to Options: A . &nbsp X B . &nbsp Y C . &nbsp N D . &nbsp None of these

Answer: Option B : B Since 4 n − 3 n − 1 = ( 3 + 1 ) n − 3 n − 1 = 3 n + n C 1 3 n − 1 + n C 2 3 n − 2 + . . . . . + n C n − 1 3 + n C n − 3 n − 1 = n C 2 3 2 + n C 3 .3 3 + . . . + n C n 3 n , ( n C 0 = n C n , n C n − 1 = n C 1 . . . . . s o o n . ) = 9 [ n C 2 + n C 3 ( 3 ) + . . . . . . + n C 4 3 n − 1 ] ∴ 4 n − 3 n − 1 is a multiple of 9 for n ≥ 2 . F o r n = 1 , 4 n − 3 n − 1 = 4 − 3 − 1 = 0 F o r n = 2 , 4 n − 3 n − 1 = 16 − 6 − 1 = 9 ∴ 4 n − 3 n − 1 is a multiple of 9 for all n ϵ N ∴ X contains elements, which are multiples of 9, and clearly Y contains all multiples of 9. ∴ X ⊂ Y i . e . , X ∪ Y = Y

If X = {4n - 3n - 1 : n ∈ N} and Y = { 9(n-1) : n ∈ N}

Question

If X = { $4^{n}$ - 3n - 1 : n ∈ N} and Y = { 9(n-1) : n ∈ N}, then X ∪ Y is equal to

Options:

A . X

B . Y

C . N

D . None of these

Answer: Option B
:
B

Since
$4^{n} - 3 n - 1 = (3 + 1)^{n} - 3 n - 1 = 3^{n} +^{n} C_{1} 3^{n - 1} +^{n} C_{2} 3^{n - 2} + . . . . . +^{n} C_{n - 1} 3 +^{n} C_{n} - 3 n - 1 =^{n} C_{2} 3^{2} +^{n} C_{3} {.3}^{3} + . . . +^{n} C_{n} 3^{n}, (^{n} C_{0} =^{n} C_{n},^{n} C_{n - 1} =^{n} C_{1} . . . . . s o o n .) = 9 [^{n} C_{2} +^{n} C_{3} (3) + . . . . . . +^{n} C_{4} 3^{n - 1}]$
$∴ 4^{n} - 3 n - 1$ is a multiple of 9 for $n \geq 2$ .
$F o r n = 1, 4^{n} - 3 n - 1 = 4 - 3 - 1 = 0 F o r n = 2, 4^{n} - 3 n - 1 = 16 - 6 - 1 = 9 ∴ 4^{n} - 3 n - 1$ is a multiple of 9 for all $n ϵ N$
$∴ X$ contains elements, which are multiples of 9, and clearly $Y$ contains all multiples of 9.
$∴ X \subset Y i . e ., X \cup Y = Y$