Question #661df

2 Answers
Aug 10, 2017

#((n+1)!)/(n+1) = n!" "# (see below, the expression changes!)

Explanation:

This expression can't really be simplified algebraically...

With that said, for fun I'll simplify the expression

#((n+1)!)/(n+1)#

#bb("METHOD 1:"#

In order to simplify the expression

#((n+1)!)/(n+1)#

one thing we could to is set up a table of values by plugging in numbers for #n#:

  • #n = 1#:

  • #"result" = 1#

  • #n = 2#:

  • #"result" = 2#

  • #n = 3#:

  • #"result" = 6#

  • #n = 4#:

  • #"result" = 24#

  • #n = 5#:

  • #"result" = 60#

You may be noticing a pattern here: these values yield the same things if the expression were #n!#:

  • #1! = 1#

  • #2! = 2#

  • #3! = 6#

  • #4! = 24#

  • #5! = 60#

Therefore,

#((n+1)!)/(n+1) = color(blue)(n!#

#" "#

#bb("METHOD 2:"#

We can also use the definition of the factorial, multiplying values decreasing by #1# each time:

#((n+1)!)/(n+1) = (cancel((n+1))(n)(n-1)(n-2)(n-3)(n-4))/(cancel(n+1))#

Notice how after the two terms are canceled, we're left with

#(n)(n-1)(n-2)(n-3)(n-4)#

Which is equivalent to #color(blue)(n!# because each successive term is one integer lower than before it:

#(n)(n-1)(n-2)(n-3)(n-4)... = color(blue)(n!#

Feb 18, 2018

It cannot be simplified.

Explanation:

I suspect this question has a typo. If the question was

#((n-1)!)/((n+1)!)#

(note the additional !), then it could be simplified.