Factorials with variables?

Question

Can someone give me a good explination of how to work with factorials with variables for example

to simplify (n+1)!/n! you get (n+1) as an answer, but how?

and how would I go about doing one like (2n+2)!/2n!

Answer 1

Notice that:
2! = 2*(1) = 2*1!
3! = 3*(2*1) = 3*2!
4! = 4*(3*2*1) = 4*3!
…
n! = n*(n - 1)!
(n + 1)! = (n + 1)*n!

A factorial can be written as the first factor times the factorial of the next factor, ie (n + 1)! = (n + 1)*n!. Applying this formula, we obtain:
(n + 1)!/n!
= (n + 1)*n!/n!
= n + 1

For (2n + 2)!/(2n)!, just apply the formula twice:
(2n + 2)!/(2n)!
= (2n + 2)*(2n + 1)!/(2n)!
= (2n + 2)*(2n + 1)*(2n)!/(2n)!
= (2n + 2)*(2n + 1)

I hope this helps!

Answer 2

(9x + 5) (4x - 3)

Answer 3

y3xy-5 multiply and simplify

Answer 4

(n+2)!-n!/(n+1)!

Answer 5

(11x plus 11) (11x plus 10)

Answer 6

multiply and simplify 7/10 X 37/150=

Answer 7

Muka mo!!!!!

Answer 8

Notice that:
2! = 2*(1) = 2*1!
3! = 3*(2*1) = 3*2!
4! = 4*(3*2*1) = 4*3!
…
n! = n*(n - 1)!
(n + 1)! = (n + 1)*n!

A factorial can be written as the first factor times the factorial of the next factor, ie (n + 1)! = (n + 1)*n!. Applying this formula, we obtain:
(n + 1)!/n!
= (n + 1)*n!/n!
= n + 1

For (2n + 2)!/(2n)!, just apply the formula twice:
(2n + 2)!/(2n)!
= (2n + 2)*(2n + 1)!/(2n)!
= (2n + 2)*(2n + 1)*(2n)!/(2n)!
= (2n + 2)*(2n + 1)

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Answer 9

Notice that:
2! = 2*(1) = 2*1!
3! = 3*(2*1) = 3*2!
4! = 4*(3*2*1) = 4*3!
…
n! = n*(n - 1)!
(n + 1)! = (n + 1)*n!

A factorial can be written as the first factor times the factorial of the next factor, ie (n + 1)! = (n + 1)*n!. Applying this formula, we obtain:
(n + 1)!/n!
= (n + 1)*n!/n!
= n + 1

For (2n + 2)!/(2n)!, just apply the formula twice:
(2n + 2)!/(2n)!
= (2n + 2)*(2n + 1)!/(2n)!
= (2n + 2)*(2n + 1)*(2n)!/(2n)!
= (2n + 2)*(2n + 1)

Algebra Homework Help - http://math.tutorcircle.com/algebra/
 

Answer 10

Multiply and simplify (a+5)(a-3)