Question

Simplify the following factorial expression: ${\displaystyle \frac{(2n+2)!}{(2n)!}}$

Answer 1

Hint:We have to simplify the factorial expression given in the fractional form. We use the definition of factorial of $n$ that is given by the formula
$n!=n\times (n-1)\times (n-2)\times ......\times 3\times 2\times 1$and use it to solve the problem.

Complete solution step by step:
Firstly we write the expression given in the question
${\displaystyle \frac{(2n+2)!}{(2n)!}}\text{ - - - - - - - - - (1)}$
Now, we use the definition of factorial to see how we can expand the factorials of both numerator and denominator i.e.
$n!=n\times (n-1)\times (n-2)\times ......\times 3\times 2\times 1$
We can see that factorial of a number is product of all integers starting from 1 to that number so we can expand the given values in our question like this
$$\begin{gathered} (2n+2)!=(2n+2)\times (2n+1)\times (2n)\times .......\times 2\times 1 \\ \text{and} \\ 2n!=2n\times (2n-1)\times (2n-2)\times ......\times 2\times 1 \end{gathered}$$
Now we put these values in the given equation (1) to see how we can proceed
${\displaystyle \frac{(2n+2)!}{(2n)!}}={\displaystyle \frac{(2n+2)\times (2n+1)\times (2n)\times .......\times 2\times 1}{2n\times (2n-1)\times ......\times 2\times 1}}$
Now we see that that both in denominators we have the expansion of $2n!$ so we write it as
$${\displaystyle \frac{(2n+2)!}{(2n)!}}={\displaystyle \frac{(2n+2)\times (2n+1)\times (2n)\times .......\times 2\times 1}{2n\times (2n-1)\times ......\times 2\times 1}}\left\{\because 2n!=2n\times (2n-1)\times ......\times 2\times 1\right\}$$
Now we can cancel the terms like this
$$\begin{gathered} {\displaystyle \frac{(2n+2)!}{(2n)!}}={\displaystyle \frac{(2n+2)\times (2n+1)\times 2n!}{2n!}} \\ \Rightarrow {\displaystyle \frac{(2n+2)!}{(2n)!}}=(2n+2)\times (2n+1) \end{gathered}$$
So we have simplified the fraction in this form.
Additional information: We can check our answer by putting a value of $n$ in the expression and solve it then we check if it is the same when the substitution is made in the result expression i.e.
Let $n=2$ so we have
$$\begin{gathered} {\displaystyle \frac{(2n+2)!}{2n!}}={\displaystyle \frac{(2\times 2+2)!}{(2\times 2)!}} \\ ={\displaystyle \frac{6!}{4!}} \\ ={\displaystyle \frac{6\times 5\times 4!}{4!}} \\ =30 \end{gathered}$$
And now putting the same value in the RHS
$$\begin{gathered} (2n+2)(2n+1)=(2\times 2+2)\times (2\times 2+1) \\ =6\times 5 \\ =30 \end{gathered}$$
So we have obtained the same answer by putting the value of ‘n’.

Note: Knowing properties of factorials helps us in this problem. Factorials can be represented in two ways by putting an exclamation mark (!) after an expression or by putting the number in an ‘L’ shaped symbol. Both the notations are correct and acceptable by the math fraternity.