Algebra problem ordinary numbers

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The discussion centers on simplifying the expression (k+1)!(k+2) - 1 to (k+2)! - 1. Participants clarify that (k + 2)! can be expressed as (k + 2) * (k + 1)!. The step-by-step breakdown shows that (k + 1)!(k + 2) equals (k + 2) * (k + 1)!. Confusion arises around the factorial notation, but it is confirmed that (k + 2)! indeed equals (k + 2) * (k + 1) * k!. The conversation emphasizes understanding factorial relationships in algebraic expressions.
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http://www.wolframalpha.com/input/?i=(k+1)!(k+2)+-1

Could anyone explain how you get from (k+1)!(k+2) - 1 to (k+2)! - 1

step by step please, here is my attempts:

(k+1)k!(k+2) - 1
(k+1)(k+2)k! - 1
(k+1)(k+2)! -1
no idea where to go from here
 
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synkk said:
http://www.wolframalpha.com/input/?i=(k+1)!(k+2)+-1

Could anyone explain how you get from (k+1)!(k+2) - 1 to (k+2)! - 1
This is very straightforward. (k + 2)! = (k + 2)*(k + 1)!
synkk said:
step by step please, here is my attempts:

(k+1)k!(k+2) - 1
(k+1)(k+2)k! - 1
(k+1)(k+2)! -1
no idea where to go from here
 
Mark44 said:
This is very straightforward. (k + 2)! = (k + 2)*(k + 1)!

how? isn't (k+2)*(k+1)k!? how is that (k+2)!?
 
synkk said:
how? isn't (k+2)*(k+1)k!?
Isn't (k+2)*(k+1)k! what?
synkk said:
how is that (k+2)!?

(k + 2)! = (k + 2)*(k + 1)! = (k + 2) * (k + 1) * k!

Let's look at ordinary numbers.

8! = 8 * 7! = 8 * 7 * 6!
= 8 * 7 * (6 * 5 * 4 * 3 * 2 * 1)
= 56 * 720 = 40320
 
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