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Question: How do you simplify \(\dfrac{3\left( n+1 \right)!}{5n!}\)?...

How do you simplify 3(n+1)!5n!\dfrac{3\left( n+1 \right)!}{5n!}?

Explanation

Solution

In this problem we need to simplify the given equation. We can observe that the given equation contains factorials function. Which is nothing but the product of the integer and all the integers below it is called a factorial of the number. In the problem we can observe the factorial function at two places: one is (n+1)!\left( n+1 \right)!, another one is n!n!. So, we will first calculate the values of both the function i.e., (n+1)!\left( n+1 \right)!, n!n!. After calculating those values, we will calculate the value of (n+1)!n!\dfrac{\left( n+1 \right)!}{n!}. Now we will multiply the calculated (n+1)!n!\dfrac{\left( n+1 \right)!}{n!} with 35\dfrac{3}{5} to get the required result.

Complete step-by-step solution:
Given that, 3(n+1)!5n!\dfrac{3\left( n+1 \right)!}{5n!}.
We can see that the factorial function is applied two times in the above equation. One is (n+1)!\left( n+1 \right)!, another one is n!n!.
From the definition of the factorial function, we can write the values of (n+1)!\left( n+1 \right)!, n!n! as
(n+1)!=(n+1)×(n+1−1)×(n+1−2)×(n+1−3)×....×3×2×1 ⇒(n+1)!=(n+1)(n)(n−1)(n−2)......3.2.1 \begin{aligned} & \left( n+1 \right)!=\left( n+1 \right)\times \left( n+1-1 \right)\times \left( n+1-2 \right)\times \left( n+1-3 \right)\times ....\times 3\times 2\times 1 \\\ & \Rightarrow \left( n+1 \right)!=\left( n+1 \right)\left( n \right)\left( n-1 \right)\left( n-2 \right)......3.2.1 \\\ \end{aligned}
n!=n(n−1)(n−2)(n−3)....3.2.1n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)....3.2.1
Now the value of (n+1)!n!\dfrac{\left( n+1 \right)!}{n!} will be calculated by taking the ratio of (n+1)!\left( n+1 \right)! and n!n!. Then we will get
(n+1)!n!=(n+1)(n)(n−1)(n−2)....3.2.1n(n−1)(n−2)....3.2.1\dfrac{\left( n+1 \right)!}{n!}=\dfrac{\left( n+1 \right)\left( n \right)\left( n-1 \right)\left( n-2 \right)....3.2.1}{n\left( n-1 \right)\left( n-2 \right)....3.2.1}
Cancelling the common terms which are in both numerator and denominator, then we will get
(n+1)!n!=n+1\dfrac{\left( n+1 \right)!}{n!}=n+1
Now multiplying the above equation with 35\dfrac{3}{5}, then we will get
35×(n+1)!n!=35×(n+1) ⇒3(n+1)!5n!=3(n+1)5 \begin{aligned} & \dfrac{3}{5}\times \dfrac{\left( n+1 \right)!}{n!}=\dfrac{3}{5}\times \left( n+1 \right) \\\ & \Rightarrow \dfrac{3\left( n+1 \right)!}{5n!}=\dfrac{3\left( n+1 \right)}{5} \\\ \end{aligned}

Note: In this problem we have calculated value of 3(n+1)!5n!\dfrac{3\left( n+1 \right)!}{5n!} by calculating the value of (n+1)!n!\dfrac{\left( n+1 \right)!}{n!} from each individual value. We have direct formula for the equation (n+1)!n!\dfrac{\left( n+1 \right)!}{n!} which is (n+1)!n!=n+1\dfrac{\left( n+1 \right)!}{n!}=n+1. We can directly use this formula to get the solution.