Question

3(n+1)! / 5n!

Please show your steps!

Answer 1

hint: \[(n+1)! \implies (n+1) \times n!\]
does that help?

Answer 2

Hint:

n! = n*(n-1)(n-2)...3*2*1

so (n+1)! = (n+1)*n*(n-1)(n-2)...3*2*1

Answer 3

and you can rewrite the remaining bit n*(n-1)(n-2)...3*2*1 as n! like lgbasallote did to get

(n+1)! = (n+1)*n!

Answer 4

the n in n! is not the same as the variable 'n' which is alone in this case?

Answer 5

not sure what you mean

Answer 6

can I cancel out the n!

Answer 7

yes