Question

If $A$ and $B$ are coefficients of ${{x}^{n}}$ in the expansion of ${{\left( 1+x \right)}^{2n}}$ and ${{\left( 1+x \right)}^{2n-1}}$ respectively, then which of the following is true?
A. $A=B$
B. $A=2B$
C. $2A=B$
D. $A+B=0$

Answer 1

Hint: First, we will be defining the terms like what are binomial expressions, what are binomial theorems and then define what are binomial coefficients. And then we will write the expressions given in the question and then find the coefficients by applying the formula for coefficients of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{m}}$ that is $^{m}{{C}_{r}}$ . And after that we will divide both the coefficients and find a relation between them and get the answer.

Complete step by step answer:
We are given binomial expressions and so for it’s understanding we must learn a few things first.
Let’s first understand a theorem called the binomial theorem: So, the Binomial Theorem is the method of expanding an expression which has been raised to any finite power. A binomial Theorem is a powerful tool of expansion, which has application in Algebra, probability, etc.
Now, let’s see the definition of a binomial expression. So, a binomial expression is an algebraic expression which contains two dissimilar terms. For example, $a+b,{{a}^{3}}+{{b}^{3}}$ etc.
If you want to expand a binomial expression with some higher power, then we use Binomial theorem formula which is as follows:
${{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{y}^{r}}}$
Here: $n\in \mathbb{N}\text{ and }x,y\in \mathbb{R}$ and $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
Now, finally let’s define the meaning of coefficients, Binomial coefficients refer to the integers which are coefficients in the binomial theorem.
Now, let’s take up are question, we are given that $A$ is coefficient of ${{x}^{n}}$ in the expansion of ${{\left( 1+x \right)}^{2n}}$ and $B$ is coefficient of ${{x}^{n}}$ in the expansion of ${{\left( 1+x \right)}^{2n-1}}$ .
Now, we know that the coefficient of ${{x}^{r}}$ in the expansion of ${{\left( 1+x \right)}^{m}}$ is $^{m}{{C}_{r}}$ , therefore:
$\begin{align}
  & \Rightarrow A{{=}^{2n}}{{C}_{n}}=\dfrac{2n!}{n!\left( 2n-n \right)!}=\dfrac{2n!}{n!n!} \\
 & \Rightarrow B{{=}^{2n-1}}{{C}_{n}}=\dfrac{\left( 2n-1 \right)!}{n!\left( 2n-1-n \right)!}=\dfrac{\left( 2n-1 \right)!}{n!\left( n-1 \right)!} \\
\end{align}$
Now, We have :
$\begin{align}
  & \Rightarrow A=\dfrac{2n!}{n!n!} \\
 & \Rightarrow B=\dfrac{\left( 2n-1 \right)!}{n!\left( n-1 \right)!} \\
\end{align}$
Now, we will divide both these terms, so we will get:
$\begin{align}
  & \Rightarrow \dfrac{A}{B}=\dfrac{\left( \dfrac{2n!}{n!n!} \right)}{\left( \dfrac{\left( 2n-1 \right)!}{n!\left( n-1 \right)!} \right)}=\dfrac{2n}{n}=2\Rightarrow \dfrac{A}{B}=2 \\
 & \Rightarrow A=2B \\
\end{align}$

So, the correct answer is “Option B”.

Note: For making the calculation easy at the final step you must know how to simplify the factorial terms like we know that we can write: $n!=n\left( n-1 \right)!$ and eventually cancel out the common terms and get the answer. Note that, the binomial coefficients which are equidistant from the beginning and from the ending are equal that is $^{n}{{C}_{0}}{{=}^{n}}{{C}_{n}}{{,}^{n}}{{C}_{1}}{{=}^{n}}{{C}_{n-1}}$ etc.