Wednesday, April 10, 2019

discrete mathematics - Prove frac1cdot3cdot5cdots(2n1)2n(n+1)!cdot4n=frac1n+12nchoosen



Prove:
1×3×5××(2n1)2n(n+1)!×4n=1n+1(2nn)





I'm having a bit of trouble proving this, I tried to prove this by induction and would get stuck. Any help is appreciated in advance.


Answer



By multiplying both numerator and denominator by 246(2n)=2nn! we get
135(2n1)2n(n+1)!=123456(2n1)2n2n(n+1)!2nn!=14n(n+1)(2n)!n!n!=14n(n+1)(2nn)



Then
4n[135(2n1)2n(n+1)!]=1n+1(2nn)


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f:AB and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...