Question

Find n, if $\left( n+1 \right)!=12\times \left( n-1 \right)!$.

Answer 1

Hint: By using the definition of factorial expand each and every term till you get common terms on both sides. Expand the $\left( n+1 \right)!$ term into 2 steps and then cancel the common terms. Now you have a quadratic equation on the left hand side and a constant on right. So by subtracting the constant on both sides we can get a quadratic equation. Then use the quadratic factorization method to solve the equation and find the value of n. The value of n is the required result in this question.

Complete step-by-step solution -
Factorial:- In mathematics, factorial is an operation, denoted by “$\left( ! \right)$ “. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as: $n!=n\times \left( n-1 \right)\times \left( n-2 \right)...................\times 1$
For example: $5!=5\times 4\times 3\times 2\times 1=120.$
Note that we assume $0!=1$ . It is standard value. It has a wide range of applications in combinations.
Given equation in the question in terms of variable n, is:
$\left( n+1 \right)!=12\times \left( n-1 \right)!$ .
Now by factorial definition, we’ll break $\left( n-1 \right)!$ by 2 steps:
$\left( n+1 \right).n.\left( n-1 \right)!=12\times \left( n-1 \right)!$ .
As we know factorials can have been zero, we cancel common terms.
$\left( n+1 \right).n.=12$
By simplifying the above equation, we can write it as:
${{n}^{2}}+n=12$
By subtraction 12 on both sides of equation, we get it as:
${{n}^{2}}+n-12=12-12$
By simplifying the above equation, we get the equation as:
${{n}^{2}}+n-12=0$
We can write the term n as $4n-3n$ , we get it as:
${{n}^{2}}+4n-3n-12=0$
By taking n common in first two terms, we get it as:
$n\left( n+4 \right)-3n-12=0$
By taking $\left( n+4 \right)$ common from whole equation, we get it as –
$\left( n+4 \right)\left( n-3 \right)=0$
By simplifying the above equation, we get its root as:
$n=3,-4$
As the question has the term $\left( n+1 \right)!$ $n+1$ must be positive.
So, n cannot be -4. So, n must be 3.
Therefore, the value of n to satisfy the given condition must be 3.

Note: In this solution we cancelled a term. We can do that step only if we know that the term is not zero. The step to avoid -4 as a result is very important. You must report only one solution. Students confuse and report both solutions. That’s why whenever you solve a question, verification is a must and compulsory step.