How do you simplify the factorial expression \[\dfrac{\left( n+1 \right)!}{n!}\]?
Answer
574.8k+ views
Hint: 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: - \[x!=x\times \left( x-1 \right)!\], 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 \[\dfrac{\left( n+1 \right)!}{n!}\] 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: -
\[\begin{align}
& \Rightarrow 5!=5\times 4\times 3\times 2\times 1 \\
& \Rightarrow 5!=120 \\
\end{align}\]
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.
Complete answer:
Here, we have been provided with the factorial expression \[\dfrac{\left( n+1 \right)!}{n!}\] 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: -
\[\begin{align}
& \Rightarrow 5!=5\times 4\times 3\times 2\times 1 \\
& \Rightarrow 5!=120 \\
\end{align}\]
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.
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