Question

How do you simplify \[\dfrac{\left( 4n+3 \right)!}{\left( 4n-1 \right)!}\] ?

Answer 1

Hint: To solve the equation we know that $n!=n\times \left( n-1 \right)!$
 Hence we can use the above formula repetitively to expand the numerator till we get the terms $\left( 4n-1 \right)!$ in the numerator. Now we will cancel the terms $\left( 4n-1 \right)!$ from the numerator and denominator and hence find the simplified expression for the given expression.

Complete step by step solution:
Now let us first understand the concept of factorial. Factorial of a number is represented by !
Factorial of a number means multiplication of all the whole numbers which are lesser than or equal to the given number. Hence let us say we want to find the factorial of a number n. Then we have $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ...\times 3\times 2\times 1$ .
Now let us understand this with an example.
Let us say we want to find $4!$ then we have $4!=4\times 3\times 2\times 1=24$ .
Hence the value of 4! Is 24.
Now by definition of factorial we can say that $n!=n\times \left( n-1 \right)!$ .
Now consider the given example \[\dfrac{\left( 4n+3 \right)!}{\left( 4n-1 \right)!}\].
Now using $n!=n\times \left( n-1 \right)!$ several times we can get,
\[\Rightarrow \dfrac{\left( 4n+3 \right)!}{\left( 4n-1 \right)!}=\dfrac{\left( 4n+3 \right)\times \left( 4n+2 \right)\times \left( 4n+1 \right)\times \left( 4n \right)\times \left( 4n-1 \right)!}{\left( 4n-1 \right)!}\]
Now canceling the term \[\left( 4n-1 \right)!\] from numerator and denominator we get,
\[\Rightarrow \dfrac{\left( 4n+3 \right)!}{\left( 4n-1 \right)!}=\left( 4n+3 \right)\times \left( 4n+2 \right)\times \left( 4n+1 \right)\times \left( 4n \right)\]
Hence can write the given expression as \[\left( 4n+3 \right)\times \left( 4n+2 \right)\times \left( 4n+1 \right)\times \left( 4n \right)\] .
Hence the simplified expression is \[\left( 4n+3 \right)\times \left( 4n+2 \right)\times \left( 4n+1 \right)\times \left( 4n \right)\]

Note:
Now we know that the factorial is defined as $n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ...\times 3\times 2\times 1$
So according to this 1! = 1. Now we have the value of 0! = 1. We cannot find the value of 0! With the help of formula as there are no whole numbers before 0. But we know that $n!=n\times \left( n-1 \right)!$
Hence substituting n = 1 we get the value of 0! Also remember the factorial is defined just for non-negative integers.