karush said:So finally
$$\frac{\left(a+1\right)!}{\left(a-2\right)!}=\frac{\left(a+1\right)a\left(a-1\right)\left(a-2\right)...}{\left(a-2\right)!}=\left(a+1\right)a\left(a-1\right)$$
karush said:how is a strike though done with latex didn't see a tool for that on the menu?
The expression $(a+1)!/(a-2)!$ is a mathematical notation that represents the ratio of the factorial of a number increased by one to the factorial of the same number decreased by two. The exclamation mark in this notation stands for the factorial function, which is the product of all positive integers less than or equal to the given number.
To simplify $(a+1)!/(a-2)!$, you can use the property of factorials that states $(n+1)! = (n+1)\times n!$. This means that $(a+1)!$ can be rewritten as $(a+1)\times a!$. Similarly, $(a-2)!$ can be rewritten as $(a-2)\times (a-3)!$. Substituting these values in the original expression, we get $(a+1)!/(a-2)! = (a+1)\times a! / [(a-2)\times (a-3)!]$. The common factor of $(a-2)$ cancels out, leaving us with the simplified form of $(a+1)a$.
When $a=2$, the expression $(a+1)!/(a-2)!$ becomes $(2+1)!/(2-2)! = 3!/0!$. Since the factorial of 0 is defined as 1, this simplifies to 3.
No, $(a+1)!/(a-2)!$ cannot be simplified further as it is already in its simplest form, $(a+1)a$.
There is no direct relationship between $(a+1)!/(a-2)!$ and $(a+1)(a-2)$. However, using the same property of factorials as mentioned in the second question, we can expand $(a+1)(a-2)$ to get $(a+1)(a-2) = (a+1)\times (a-2)\times (a-3)!$. We can then cancel out the common factors of $(a+1)$ and $(a-2)$ to get the simplified form of $(a+1)!/(a-2)!$, which is $(a+1)a$.