Reducing Factorials: Is the Answer 1/[2n*2n+1]?

AI Thread Summary
The discussion centers on simplifying the expression (2n-1)!/(2n+1)!. The solution involves recognizing that this can be reduced to 1/[2n(2n+1)]. The participant suggests that adding parentheses improves clarity. The final answer is confirmed as correct, emphasizing the importance of proper notation in mathematical expressions. The conversation highlights the significance of clear communication in solving factorial problems.
pugfug90
Messages
118
Reaction score
0

Homework Statement


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


Homework Equations





The Attempt at a Solution



...2n-2*2n-1
------------ (pretend that's a divider)
...2n-2*2n-1*2n*2n+1

Is the answer 1/[2n*2n+1]?
 
Physics news on Phys.org
It sure is.
 
But it would look better with an additional pair of parentheses:
1/[2n(2n+1)]. I would have been inclined to interpret what you have as
1/(2n2+1).
 
Thanks :)
 
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
Back
Top