
If the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n}}\] is ‘\[a\]’ and the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n - 1}}\] is ‘\[b\]’, then \[\dfrac{a}{b} = \]
A) 2
B) 4
C) \[2n\]
D) \[n\]
Answer
558k+ views
Hint:
Here we will first write the general form of the standard equation i.e. \[{\left( {p + q} \right)^n}\]. Then by using this, we will find the value of the coefficient of \[{x^n}\] in the given equations. Then we will divide the coefficients of \[{x^n}\] to get the required ratio i.e. \[\dfrac{a}{b}\].
Complete step by step solution:
First, we will write the general term of the \[{\left( {p + q} \right)^n}\].
The general term of binomial expansion is \[T = {}^n{C_r} \cdot {p^{n - r}} \cdot {q^r}\].
Now we will find the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n}}\] by using the above general term.
Therefore, we put \[p = 1,q = x,n = 2n\] and \[r = n\] in the general term equation. Therefore, we get
\[T = {}^{2n}{C_n} \cdot {1^{2n - n}} \cdot {x^n}\]
Simplifying the expression, we get
\[ \Rightarrow T = {}^{2n}{C_n} \cdot {x^n}\]…………………………\[\left( 1 \right)\]
It is given that the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n}}\] is ‘\[a\]’ .
Now from equation \[\left( 1 \right)\], we can see that the coefficient of \[{x^n}\] is \[{}^{2n}{C_n}\]. So,
\[a = {}^{2n}{C_n}\]
Now we will find the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n - 1}}\] by using the above general term.
Therefore, we put \[p = 1,q = x,n = 2n - 1\] and \[r = n\] in the general term equation. Therefore, we get
\[T = {}^{2n - 1}{C_n} \cdot {1^{2n - 1 - n}} \cdot {x^n}\]
Simplifying the expression, we get
\[ \Rightarrow T = {}^{2n - 1}{C_n} \cdot {x^n}\]………………………\[\left( 2 \right)\]
It is given that the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n}}\] is ‘\[b\]’.
Now from equation \[\left( 2 \right)\], we can see that the coefficient of \[{x^n}\] is \[{}^{2n - 1}{C_n}\]. So,
\[b = {}^{2n - 1}{C_n}\]
Now we will find the ratio of \[a\] to \[b\]. Therefore, we get
\[\dfrac{a}{b} = \dfrac{{{}^{2n}{C_n}}}{{{}^{2n - 1}{C_n}}}\]
Now we will expand this combination form using the formula \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\] to get the required ratio. Therefore, we get
\[ \Rightarrow \dfrac{a}{b} = \dfrac{{\dfrac{{2n!}}{{n!\left( {2n - n} \right)!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{n!\left( {2n - 1 - n} \right)!}}}}\]
Subtracting the terms, we get
\[ \Rightarrow \dfrac{a}{b} = \dfrac{{\dfrac{{2n!}}{{n!n!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{n!\left( {n - 1} \right)!}}}}\]
Now we will solve this factorial term. Therefore, we get
\[ \Rightarrow \dfrac{a}{b} = \dfrac{{\dfrac{{2n \times \left( {2n - 1} \right)!}}{{n! \times n \times \left( {n - 1} \right)!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{n!\left( {n - 1} \right)!}}}}\]
Now we will cancel out the common terms present in the numerator and the denominator. Therefore, we get
\[ \Rightarrow \dfrac{a}{b} = \dfrac{{2n}}{n} = 2\]
Hence, the value of the ratio \[\dfrac{a}{b}\] is equal to 2.
So, option A is the correct option.
Note:
Factorial of a number is equal to the multiplication or product of all the positive integers smaller than or equal to the number. In addition, factorial of 1 is always equals to 1 and factorial of 0 is equal to 1. The factorial of a number is always positive, it can never be negative and the factorial of a negative number is not defined.
Example of factorial: \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]
Here we will first write the general form of the standard equation i.e. \[{\left( {p + q} \right)^n}\]. Then by using this, we will find the value of the coefficient of \[{x^n}\] in the given equations. Then we will divide the coefficients of \[{x^n}\] to get the required ratio i.e. \[\dfrac{a}{b}\].
Complete step by step solution:
First, we will write the general term of the \[{\left( {p + q} \right)^n}\].
The general term of binomial expansion is \[T = {}^n{C_r} \cdot {p^{n - r}} \cdot {q^r}\].
Now we will find the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n}}\] by using the above general term.
Therefore, we put \[p = 1,q = x,n = 2n\] and \[r = n\] in the general term equation. Therefore, we get
\[T = {}^{2n}{C_n} \cdot {1^{2n - n}} \cdot {x^n}\]
Simplifying the expression, we get
\[ \Rightarrow T = {}^{2n}{C_n} \cdot {x^n}\]…………………………\[\left( 1 \right)\]
It is given that the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n}}\] is ‘\[a\]’ .
Now from equation \[\left( 1 \right)\], we can see that the coefficient of \[{x^n}\] is \[{}^{2n}{C_n}\]. So,
\[a = {}^{2n}{C_n}\]
Now we will find the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n - 1}}\] by using the above general term.
Therefore, we put \[p = 1,q = x,n = 2n - 1\] and \[r = n\] in the general term equation. Therefore, we get
\[T = {}^{2n - 1}{C_n} \cdot {1^{2n - 1 - n}} \cdot {x^n}\]
Simplifying the expression, we get
\[ \Rightarrow T = {}^{2n - 1}{C_n} \cdot {x^n}\]………………………\[\left( 2 \right)\]
It is given that the coefficient of \[{x^n}\] in \[{\left( {1 + x} \right)^{2n}}\] is ‘\[b\]’.
Now from equation \[\left( 2 \right)\], we can see that the coefficient of \[{x^n}\] is \[{}^{2n - 1}{C_n}\]. So,
\[b = {}^{2n - 1}{C_n}\]
Now we will find the ratio of \[a\] to \[b\]. Therefore, we get
\[\dfrac{a}{b} = \dfrac{{{}^{2n}{C_n}}}{{{}^{2n - 1}{C_n}}}\]
Now we will expand this combination form using the formula \[^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\] to get the required ratio. Therefore, we get
\[ \Rightarrow \dfrac{a}{b} = \dfrac{{\dfrac{{2n!}}{{n!\left( {2n - n} \right)!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{n!\left( {2n - 1 - n} \right)!}}}}\]
Subtracting the terms, we get
\[ \Rightarrow \dfrac{a}{b} = \dfrac{{\dfrac{{2n!}}{{n!n!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{n!\left( {n - 1} \right)!}}}}\]
Now we will solve this factorial term. Therefore, we get
\[ \Rightarrow \dfrac{a}{b} = \dfrac{{\dfrac{{2n \times \left( {2n - 1} \right)!}}{{n! \times n \times \left( {n - 1} \right)!}}}}{{\dfrac{{\left( {2n - 1} \right)!}}{{n!\left( {n - 1} \right)!}}}}\]
Now we will cancel out the common terms present in the numerator and the denominator. Therefore, we get
\[ \Rightarrow \dfrac{a}{b} = \dfrac{{2n}}{n} = 2\]
Hence, the value of the ratio \[\dfrac{a}{b}\] is equal to 2.
So, option A is the correct option.
Note:
Factorial of a number is equal to the multiplication or product of all the positive integers smaller than or equal to the number. In addition, factorial of 1 is always equals to 1 and factorial of 0 is equal to 1. The factorial of a number is always positive, it can never be negative and the factorial of a negative number is not defined.
Example of factorial: \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

