If \[X = \left\{ {{4^n} - 3n - 1:n \in N} \right\}\]and \[Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\}\], then \[X \cup Y\] is equal to
A. \[X\]
B. \[Y\]
C. \[N\]
D. None of the above
Answer
362.4k+ views
Hint: Convert the set of elements of \[X\] in terms of the set of elements of \[Y\]. While converting the set of elements use Binomial Theorem for expanding the terms. So, use this concept to reach the solution of the problem.
Complete step-by-step answer:
Given set \[Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\}\] and
Set \[X\] contains elements of the form
\[ \Rightarrow {4^n} - 3n - 1\]
Which can be written as
\[ \Rightarrow {\left( {1 + 3} \right)^n} - 3n - 1\]
Opening the terms in the bracket by using the formula \[{\left( {1 + x} \right)^n} = {}^n{C_0} + x{}^n{C_1} + {x^2}{}^n{C_2} + ....................... + {x^{n - 1}}{}^n{C_{n - 1}} + {x^n}{}^n{C_n}\] we have,
\[
\Rightarrow 1 + 3{}^n{C_1} + {3^2}{}^n{C_2} + .............. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n} - 3n - 1 \\
\Rightarrow 1 + 3n + {3^2}{}^n{C_2} + .................. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n} - 3n - 1 \\
\]
Cancelling the common terms, we get
\[ \Rightarrow {3^2}{}^n{C_2} + .................. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n}\]
Taking \[{3^2}\]as common, we get
\[
\Rightarrow {3^2}\left( {{}^n{C_2} + ................ + {3^{n - 3}}{}^n{C_{n - 1}} + {3^{n - 2}}{}^n{C_n}} \right) \\
\Rightarrow 9\left( {{}^n{C_2} + ................ + {3^{n - 3}}{}^n{C_{n - 1}} + {3^{n - 2}}{}^n{C_n}} \right) \\
\]
Clearly, set \[X\] has natural numbers which are multiples of 9 (not all) and the set \[Y\]has all the multiples of 9.
Therefore, \[X \subset Y\]. So \[X \cup Y\] is equal to the set of elements in \[Y\].
Thus, the correct option is B. \[Y\]
Note: In the given problem the set \[X \cup Y\] has the elements of both elements of the sets\[X\] and \[Y\]. But the elements of set \[Y\] contain the elements of set \[X\] i.e., \[X \subset Y\] from the solution.
Complete step-by-step answer:
Given set \[Y = \left\{ {9\left( {n - 1} \right):n \in N} \right\}\] and
Set \[X\] contains elements of the form
\[ \Rightarrow {4^n} - 3n - 1\]
Which can be written as
\[ \Rightarrow {\left( {1 + 3} \right)^n} - 3n - 1\]
Opening the terms in the bracket by using the formula \[{\left( {1 + x} \right)^n} = {}^n{C_0} + x{}^n{C_1} + {x^2}{}^n{C_2} + ....................... + {x^{n - 1}}{}^n{C_{n - 1}} + {x^n}{}^n{C_n}\] we have,
\[
\Rightarrow 1 + 3{}^n{C_1} + {3^2}{}^n{C_2} + .............. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n} - 3n - 1 \\
\Rightarrow 1 + 3n + {3^2}{}^n{C_2} + .................. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n} - 3n - 1 \\
\]
Cancelling the common terms, we get
\[ \Rightarrow {3^2}{}^n{C_2} + .................. + {3^{n - 1}}{}^n{C_{n - 1}} + {3^n}{}^n{C_n}\]
Taking \[{3^2}\]as common, we get
\[
\Rightarrow {3^2}\left( {{}^n{C_2} + ................ + {3^{n - 3}}{}^n{C_{n - 1}} + {3^{n - 2}}{}^n{C_n}} \right) \\
\Rightarrow 9\left( {{}^n{C_2} + ................ + {3^{n - 3}}{}^n{C_{n - 1}} + {3^{n - 2}}{}^n{C_n}} \right) \\
\]
Clearly, set \[X\] has natural numbers which are multiples of 9 (not all) and the set \[Y\]has all the multiples of 9.
Therefore, \[X \subset Y\]. So \[X \cup Y\] is equal to the set of elements in \[Y\].
Thus, the correct option is B. \[Y\]
Note: In the given problem the set \[X \cup Y\] has the elements of both elements of the sets\[X\] and \[Y\]. But the elements of set \[Y\] contain the elements of set \[X\] i.e., \[X \subset Y\] from the solution.
Last updated date: 02nd Oct 2023
•
Total views: 362.4k
•
Views today: 6.62k
Recently Updated Pages
What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

An alternating current can be produced by A a transformer class 12 physics CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers
