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But for (2n)! It can only be simplified a little. We can only find factorial of whole numbers. 2n is an algebraic number.

Формула факториала n!=1⋅2⋅3⋅...⋅(n−2)⋅(n−1)⋅n.

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I'm assuming you just left off the O() notation on the left side. O(2^(n+1)) is the same as O(2 * 2^n) However, constant factors are the only thing you can pull out. 2^(2n) can be expressed as (2^n)(2^n)...

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So when dividing $(2n+2)!$ by $(2n)!$ only those first two factors of $(2n+2)!$ remain (in this case in the denominator). (N + 2)(N + 1) N!$ will help simplify the desired quotient.

Азот. 3N2 + 2e- → 2N3 N2 + 4H2O + 2e- → 2NH2OH + 2OH

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