Question

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 1

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: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$