Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

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
VerifiedVerified
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: \[5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]