2015年12月10日 · While there isn't a simplification of ((2n)!)/(n!), there are other ways of expressing it. For example ((2n)!)/(n!) = prod_(k=0)^(n-1)(2n-k) = (2n)(2n-1)...(n+1) This …

2015年11月24日 · 33201 views around the world You can reuse this answer ...

2018年3月26日 · n=7 7-2n = n-14 The first step is to move all of your variables, in this case n, to one side of the equation, and all constants (numbers without variables) to the other side: 7-2n …

2016年7月2日 · (2n!) / (n!) = prod_{k=n+1} ^{k=2n}k Suppose that we need (8!)/(4!) = (1 cdot 2 cdot 3 cdot 4 cdot 5 cdot ...

2017年2月8日 · ((2n-1)!)/((2n+1)!) = 1/((2n+1)(2n)) Remember that: n! =n(n-1)(n-2)...1 And so (2n+1)! =(2n+1)(2n)(2n-1)(2n-2) ... 1 ...

2017年2月8日 · The series is absolutely convergent for every x in RR and its sum is: sum_(n=0)^oo x^(2n)/(n!) = e^(x^2) Use the ratio test stating that a series: sum_(n=0)^oo a_n …

2018年4月20日 · Diverges by the Ratio Test. We'll check for Absolute Convergence using the Ratio Test, where a_n=(-5)^(2n)/(n^2 9^n). The Ratio Test tells us we take L=lim_(n …

2018年1月30日 · See a solution process below: To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

2018年3月23日 · #using the Stirling asymptotic formula. #n! approx sqrt(2pi n)(n/e)^n# #(((2n)!)/((n!)n^n))^(1/n) approx 1/n (sqrt(2pi(2n))/sqrt(2pi n))^(1/n) ((2n)/e)^2/(n/e) = 1/n ...

2017年9月5日 · #-n^2" is a "color(blue)"common factor"" in all 3 terms"# #rArr-n^2(n^2+3+2n)# #=-n^2(n^2+2n+3)# #"check the "color(blue)"discriminant"" of "n^2+2n+3#