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Question: How do you simplify the factorial expression \[\dfrac{\left( n+1 \right)!}{n!}\]?...

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

Explanation

Solution

First understand the meaning of ‘factorial of a number’ and the type of number for which it is defined. Consider ‘n’ as any positive integer and use the property of factorial of a number given as: - x!=x×(x1)!x!=x\times \left( x-1 \right)!, to simplify the numerator. Leave the denominator as it is and cancel the common factors to get the answer.

Complete answer:
Here, we have been provided with the factorial expression (n+1)!n!\dfrac{\left( n+1 \right)!}{n!} and we have been asked to simplify it. But first we need to know about the term ‘factorial’.
Now, in mathematics, the factorial of any number (which must be a positive integer) is the product of that number and all the positive integers less than that number. It is generally denoted by the ‘!’ sign. Let us consider an example: - here we are considering the positive integer 5 and we have to find its factorial. It can be therefore written as: -

& \Rightarrow 5!=5\times 4\times 3\times 2\times 1 \\\ & \Rightarrow 5!=120 \\\ \end{aligned}$$ Similarly, we can find the factorial of any positive integer. Here, in the above expression of factorial you may notice an interesting observation. We can write $$4\times 3\times 2\times 1=4!$$. So, we can also write the expression of 5! as: - $$\Rightarrow 5!=5\times 4!$$ In general, for a positive integer x, its factorial can be written as: - $$\Rightarrow x!=x.\left( x-1 \right).\left( x-2 \right).....1$$ Leaving x alone while grouping all other products, we get, $$\Rightarrow x!=x.\left[ \left( x-1 \right).\left( x-2 \right).\left( x-3 \right).....1 \right]$$ $$\Rightarrow x!=x.\left( x-1 \right)!$$ - (1) Now, let us come to the question. Assuming the value of the given factorial expression as ‘E’, we have, $$\Rightarrow E=\dfrac{\left( n+1 \right)!}{n!}$$ Leaving the denominator as it is while using equation (1) to simplify the numerator, we get, $$\Rightarrow E=\dfrac{\left( n+1 \right)!.n!}{n!}$$ Cancelling the common factors, we get, $$\Rightarrow E=\left( n+1 \right)$$ Hence, the value of the given factorial expression is $$\left( n+1 \right)$$. **Note:** One may note that we can check our answer easily by assigning some positive integral value to n. For example: - you can assume m = 6 and then substitute in the given expression. If you will get the answer equal to 7 then the expression that we have obtained E = (n + 1) is proved to be correct. Always remember that factorial of a number is undefined for negative integers, rational and irrational numbers etc. You may remember an important result given as 0! = 1.