How do you simplify #((4n+3)!)/((4n-1)!)#?

2 Answers
Cesareo R.
May 26, 2016

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

Explanation:

#(4n+3)! = 1 times 2 times 3 times ... times(4n-1)times(4n)times(4n+1)times(4n+2)times(4n+3)#
#(4n+3)! =(4n-1)! times(4n)times(4n+1)times(4n+2)times(4n+3)#
then
#((4n+3)! )/((4n-1)!) = (4n)times(4n+1)times(4n+2)times(4n+3)#

George C.
May 26, 2016

#((4n+3)!)/((4n-1)!)=(4n+3)(4n+2)(4n+1)(4n+0)#

#=256n^4+384n^3+176n^2+24n#

Explanation:

If #n > 0# then:

#((4n+3)!)/((4n-1)!)#

#=((4n+3)(4n+2)(4n+1)(4n+0)(color(red)(cancel(color(black)((4n-1)!)))))/(color(red)(cancel(color(black)((4n-1)!))))#

#=(4n+3)(4n+2)(4n+1)(4n+0)#

#=(16n^2+20n+6)(16n^2+4n)#

#=256n^4+384n^3+176n^2+24n#

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