NIMCET 2009 Question Paper with solutions

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Question : 20 of 120

Marks: +1, -0

If

P=\{(4^{n}-3 n-1) ∣ n ∈ N\}

and

Q=\{(9 n-9) ∣ n ∈ N\}

, then

P ∪ Q

is equal to

N
P
Q
None of these

Solution:

S=\{(4^{n}-3 n-1) ∣ n ∈ N\}

4^{n}-3 n-1=(3+1)^{n}-3 n-1

=1+3 n+{ }^{n} C_{2}-3^{2}+{ }^{n} C_{3} ⋅ 3^{3}+. . . .+{ }^{n} C_{n} ⋅ 3 n-3 n-1

=9 {F}^{n} C_{2}+3^{n} C_{3}+. . .]

=M(9)

Q=\{(9 n-9) ∣ n ∈ N\}=\{0,9,18 . . .\}

includes all non-negative integers divisible by 9 .

⇒ \;\; P ⊆ Q ⇒ P ∪ Q=Q

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