SOLUTION: Show that:
2[(2n-1)C(n-1)] = 2nCn
Algebra.Com
Question 920296: Show that:
2[(2n-1)C(n-1)] = 2nCn
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
!})
!}{r!*((2n-1)-r)!})
!}{(n-1)!*((2n-1)-(n-1))!})
!}{(n-1)!*(2n-1-n+1)!})
!}{(n-1)!*(n)!})
!}{2n(n-1)!*(n)!})
Multiply top and bottom by 2n.
!}{2*(n(n-1)!)*(n)!})
!}{2*(n)!*(n)!})
!}{n!*n!})
!}{n!*(2n-n)!})
)
 = \ \ \ _{2n}C_{n})
RELATED QUESTIONS
Show that:
n(2nCn) = (n+1)[2nC(n+1)] (answered by richard1234)
Prove that 2C(2n − 1,n) = C(2n,n) for all n >... (answered by greenestamps)
Hi
How exactly do I do b + c of this question?
a) Using a large scale, sketch the... (answered by stanbon)
How does 2n(n+1)-2n^2 simplify to... (answered by Fombitz)
{{{ 1/ (2n^2) }}} + {{{ 5 / 2n }}} = {{{ (n-2) / n^2... (answered by ankor@dixie-net.com)
Show that in any GP, S(2n): S(n)=(r^n+1):1
Thanks! (answered by Edwin McCravy)
n^3-2n^2(n+1) (answered by MathLover1)
2+4+6+.....2n=n(n+1) (answered by ikleyn)
(n)^2+(2n-1)^2=(2n+1)^2 (answered by checkley79)