Question

Prove that $\dfrac{\left( 2n \right)!}{n!}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}.$

Answer 1

Hint: The factorial of a natural number is defined as the product of all the natural numbers less than or equal to that number. That is the factorial of a natural number $n$ is the product of the natural numbers from $1$ to $n.$ So, $n!=1\cdot 2\cdot 3\cdot ...\cdot \left( n-1 \right)n.$

Complete step by step answer:
Consider the given problem $\dfrac{\left( 2n \right)!}{n!}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}.$
We have already learnt that the factorial of a natural number is defined as the product of all the natural numbers less than or equal to that number.
Suppose we need to find the factorial of a natural number $n.$ So, we will get the identity $n!=1\cdot 2\cdot 3\cdot ...\cdot \left( n-1 \right)n.$
Let us consider the right-hand side of the given equation, $1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}.$
Here, the product of the odd numbers from $1$ to $2n-1$ is multiplied to ${{2}^{n}}.$
Now, we need to make this a fraction in terms of factorials. In order to make this in terms of factorial we are going to multiply and divide it with the product $2\cdot 4\cdot 6\cdot ...\cdot 2n.$
So, we will get $1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}\dfrac{2\cdot 4\cdot 6\cdot ...\cdot 2n}{2\cdot 4\cdot 6\cdot ...\cdot 2n}.$
We know that $1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}\cdot 2\cdot 4\cdot ...\cdot 2n=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}.$
So, we will get $1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}=\dfrac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}}{2\cdot 4\cdot ...\cdot 2n}.$
By the definition of the factorial, we will get the following $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n=2n!.$
Therefore, we will get the numerator of the above fraction in terms of factorial.
That is, $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}=\dfrac{\left( 2n \right)!{{2}^{n}}}{2\cdot 4\cdot ...\cdot 2n}.$
Let us consider the denominator of the above fraction, $2\cdot 4\cdot ...\cdot 2n.$ Since all the terms are even, all of them have a common factor. That is, $2.$ If we are taking $2$ from each of these numbers, we will get $2\cdot 4\cdot 6\cdot ...\cdot 2n=2\cdot 2\cdot 2\cdot ...\cdot {{2}_{n times}}\cdot 1\cdot 2\cdot 3\cdot ...\cdot n.$
Thus, we will get $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}=\dfrac{\left( 2n \right)!{{2}^{n}}}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n\cdot {{2}^{n}}}.$
Let us cancel the common factor from the numerator and the denominator to get $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}=\dfrac{\left( 2n \right)!}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}.$
We have $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n=n!.$
Therefore, $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot 2n\cdot {{2}^{n}}=\dfrac{\left( 2n \right)!}{n!}=\text{LHS}.$
Hence it is proved that $\dfrac{\left( 2n \right)!}{n!}=1\cdot 3\cdot 5\cdot ...\cdot \left( 2n-1 \right){{2}^{n}}.$

Note: Remember that $0!=1.$ Also, note that \[\left( 2n \right)!\ne 2n!.\] We know that if we multiply a number multiple times, then that is equal to the number raised to the number of the times it is multiplied.