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Mathematics 13 Online
OpenStudy (anonymous):

how do you get (n+1) from (n+1)!/(n)! ?

OpenStudy (anonymous):

I think it unravels like this [(n+1)(n)]/[n}, so it gives (n+1)

OpenStudy (anonymous):

because you n's cancel out

OpenStudy (anonymous):

*the

OpenStudy (anonymous):

right!

OpenStudy (anonymous):

does that help?

OpenStudy (anonymous):

\[(n+1)!/n!= (n+1)(n)(n-1)(n-2)......./n(n-1)(n-2)......\] sine the numerator and denominator both have (n)(n-1)(n-2)..../(n)(n-1)(n-2)..... they cancel out leaving n+1

OpenStudy (anonymous):

yes, it makes a lot of sense. thank you!

OpenStudy (anonymous):

that was rather simple. thanks guys

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