Question

The value of \[{2^n}\left\{ {1 \cdot 3 \cdot 5...\,\,...\left( {2n - 3} \right)\left( {2n - 1} \right)} \right\}\] is?
A. \[\dfrac{{\left( {2n} \right)!}}{{n!}}\]
B. \[\dfrac{{\left( {2n} \right)!}}{{{2^n}}}\]
C. \[\dfrac{{n!}}{{\left( {2n} \right)!}}\]
D. None of these

Answer 1

Hint: Here in this question given a series of numbers, we have to find their exact value or simplest form of the series. For this, first we need to multiply and divide the even integers and further simplify using the factorial concepts and by using basic arithmetic operations to get the required solution.

Complete step by step answer:
A number of things or events of the same class coming one after another in spatial or temporal succession is known as series.
Now consider the given series:
\[ \Rightarrow {2^n}\left\{ {1 \cdot 3 \cdot 5...\,\,...\left( {2n - 3} \right)\left( {2n - 1} \right)} \right\}\]
Multiply and divide the given series with series of even integer or Multiply and divide by \[2 \cdot 4 \cdot 6 \ldots \,\,...\left( {2n - 2} \right) \cdot 2n\], then we have
\[ \Rightarrow {2^n}\left\{ {1 \cdot 3 \cdot 5...\,\,...\left( {2n - 3} \right)\left( {2n - 1} \right)} \right\} \times \dfrac{{2 \cdot 4 \cdot 6 \ldots \,\,...\left( {2n - 2} \right) \cdot 2n}}{{2 \cdot 4 \cdot 6 \ldots \,\,...\left( {2n - 2} \right) \cdot 2n}}\]
On Multiplication, we have
\[ \Rightarrow \dfrac{{{2^n}\left\{ {1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6...\,\,...\left( {2n - 3} \right) \cdot \left( {2n - 2} \right) \cdot \left( {2n - 1} \right) \cdot 2n} \right\}}}{{2 \cdot 4 \cdot 6 \ldots \,\,...\left( {2n - 2} \right) \cdot 2n}}\]
In numerator \[1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6...\,\,...\left( {2n - 3} \right) \cdot \left( {2n - 2} \right) \cdot \left( {2n - 1} \right) \cdot 2n\] is equal to the value of \[\left( {2n} \right)!\], then the above inequality becomes
\[ \Rightarrow \dfrac{{{2^n}\left\{ {\left( {2n} \right)!} \right\}}}{{2 \cdot 4 \cdot 6 \ldots \,\,...\left( {2n - 2} \right) \cdot 2n}}\]
Take 2 as common in denominator, then
\[ \Rightarrow \dfrac{{{2^n}\left\{ {\left( {2n} \right)!} \right\}}}{{{2^n}\left( {1 \cdot 2 \cdot 3 \cdot 4 \ldots \,\,...\left( {n - 1} \right) \cdot n} \right)}}\]
On cancelling the like terms \[{2^n}\] in both numerator and denominator, then we get
\[ \Rightarrow \dfrac{{\left( {2n} \right)!}}{{1 \cdot 2 \cdot 3 \cdot 4 \ldots \,\,...\left( {n - 1} \right) \cdot n}}\]
In denominator \[1 \cdot 2 \cdot 3 \cdot 4..\,\,...\left( {n - 1} \right) \cdot n\] is equal to the value of \[n!\], then the above inequality becomes
\[\therefore \,\,\,\,\dfrac{{\left( {2n} \right)!}}{{n!}}\]
Hence, the required value of \[{2^n}\left\{ {1 \cdot 3 \cdot 5...\,\,...\left( {2n - 3} \right)\left( {2n - 1} \right)} \right\} = \dfrac{{\left( {2n} \right)!}}{{n!}}\].

So, the correct answer is “Option A”.

Note: Remember, the series of even integer means the multiple of even numbers up to ‘n’ terms i.e., \[2 \cdot 4 \cdot 6 \cdot 8...\,\,...\left( {2n - 2} \right)2n\].
The series of odd integer means the multiple of odd numbers up to the ‘n’ terms i.e., \[1 \cdot 3 \cdot 5...\,\,...\left( {2n - 3} \right)\left( {2n - 1} \right)\].
Factorial is the continued product of first ‘n’ natural numbers is called the “n factorial” and it represented by \[n! = n \cdot \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right) \cdot .....3 \cdot 2 \cdot 1\].