How do you simplify the factorial expression $ \dfrac{\left( 3n+1 \right)!}{\left( 3n \right)!} $ ?

Hint: In this question, we are given a expression in terms of factorial and we need to simplify it. For this, we will use the definition of factorial to expand our numerator upto the point where a term (factor) matches with the denominator. Then we will cancel out the terms and conclude our simplified answer. Factorial for an integral n is denoted by n! and is given as $ n!=n\cdot \left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\ldots \ldots \ldots \left( 3 \right)\left( 2 \right)\left( 1 \right) $ .

Complete step by step answer:
Here we are given an expression as $ \dfrac{\left( 3n+1 \right)!}{\left( 3n \right)!} $ . Here '!' sign denotes the factorial. So let us first understand the meaning of factorial.
Factorial is a multiplication operation of natural numbers with all natural numbers that are less than 1. For a natural number n, the factorial of it will be denoted by n! and it will be equal to $ n!=n\cdot \left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\ldots \ldots \ldots \left( 3 \right)\left( 2 \right)\left( 1 \right) $ .
Let us observe the pattern, for n! We have $ n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times \ldots \ldots \ldots \times \left( 3 \right)\times \left( 2 \right)\times \left( 1 \right) $ .
As we can see, if we take all terms except n then we have $ \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times \ldots \ldots \ldots \times \left( 3 \right)\times \left( 2 \right)\times \left( 1 \right) $ which will be equal to (n-!) so n! can also be written as n! = n(n-1)!
Let us look at our given expression $ \dfrac{\left( 3n+1 \right)!}{\left( 3n \right)!} $ .
In the numerator we have (3n+1)! It can be expanded as follows,
 $ \left( 3n+1 \right)!=\left( 3n+1 \right)\times \left( 3n \right)\times \left( 3n-1 \right)\times \left( 3n-2 \right)\times \ldots \ldots \ldots \times 3\times 2\times 1 $ .
If we take all the terms except we have $ \left( 3n \right)\times \left( 3n-1 \right)\times \left( 3n-2 \right)\times \ldots \ldots \ldots \times 3\times 2\times 1 $ .
Since 3n has all of its lower natural numbers multiplied to it, we can say it is equal to (3n)!
Therefore we can write (3n+1)! as $ \left( 3n+1 \right)!=\left( 3n+1 \right)\left( 3n \right)! $ .
Our expression therefore becomes $ \dfrac{\left( 3n+1 \right)!}{\left( 3n \right)!}=\dfrac{\left( 3n+1 \right)\left( 3n \right)!}{\left( 3n \right)!} $ .
As we can see (3n)! is multiplied in both the numerator and the denominator so we can cancel it. Therefore we are left with (3n+1) only. So we have $ \dfrac{\left( 3n+1 \right)!}{\left( 3n \right)!}=3n+1 $ which is our required simplified answer.

Note:
 Students should be careful while multiplying the lower values. For easier calculations, just keep on subtracting one from the previous factor until you reach 1. We only multiply natural numbers, so 0 will not be multiplied as it will make the factorial as 0. Students should note that 0! is always equal to 1. Factorial can also be denoted by L i.e. $ \left| \!{\underline {\,
  n \,}} \right. =n! $.