Simplifying Factorials: How to Simplify 2n+2!?

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The discussion focuses on simplifying the expression for factorials, specifically 2n + 2!. Clarification is provided that 2n + 2! equals 2n + 2, while the intended expression is (2n + 2)!, which expands to (2n + 2)*(2n + 1)*(2n)!. Participants emphasize the importance of using parentheses to avoid confusion between 2n! and (2n)!. The conversation highlights common mistakes in factorial notation and the correct simplification process. Understanding these distinctions is crucial for accurate mathematical communication.
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Just a quick question. I'm having problems with factorials. For example, how do you simplify 2n+2!? Is it 2n!*2n+1!?

Thanks everybody!
 
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Strictly speaking, 2n + 2! would be just 2n + 2, since 2! = 2. You probably meant (2n + 2)!, which is (2n + 2)*(2n + 1)*(2n)!.

Edit: added a missing factor in the above.


Note that 2n! is different from (2n)!. The first is 2(n*(n-1)*(n-2)*...*2*1). The second is 2n*(2n-1)*(2n-2)*...*2*1.

You should get into the habit of using parentheses...
 
Last edited:


Mark44 said:
You probably meant (2n + 2)!, which is (2n + 1)*(2n)!.
Actually, it's (2n+2)*(2n+1)*(2n)!, but I'm sure you knew that.
 


Yeah, knew that, but didn't manage to write it.
I edited my earlier post.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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