Calculating (n+2)! quickly and accurately - Help Needed!

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To calculate (n+2)!, start with the expansion of n! as (1)(2)(3)...(n). To find (n+2)!, add the terms (n+1) and (n+2) to the expansion of n!, resulting in (n+2)! = (1)(2)(3)...(n)(n+1)(n+2). When dividing (n+2)! by n!, the n! terms cancel, simplifying the expression to (n+2)(n+1). This method allows for quick and accurate calculation of factorials and their quotients.
Phyzwizz
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I know that n!=(1)(2)(3)...(n-1)(n)
I am confused how I can figure out what (n+2)! is in order to divide it by n!
How can I figure (n+2)! out and what is it, so that I won't have to ask if what I got is right.

Thanks!
 
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You have already written out the expansion for n!. Now, if you add 1 to n, what additional term must be included in n! to make it (n+1)! Once you figure this out, repeat to find
(n+2)!
 
3!=3*2*1
4!=4*3*2*1
10!=10*9*8*...*2*1
etc.
What is 10! divided by 8! then?
 
oh okay so (n+2)! would be (1)(2)(3)...(n-1)(n)(n+1)(n+2)?
 
Right. Now figure your quotient (n+2)!/n!.
 
Phyzwizz said:
oh okay so (n+2)! would be (1)(2)(3)...(n-1)(n)(n+1)(n+2)?
Which is the same as (n + 2)(n + 1) n!, right?
 
There is no n! in the quotient of (n+2)!/n!. Remember the expansion of (n+2)! is
(n+2)(n+1)n!. Dividing (n+2)! by n! is (n+2)(n+1)n!/n! = (n+2)(n+1).
 
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