Answer:
therefore, [tex]\frac{(2n)!}{n!}=(n+1)\cdot (n+2)\dots (2n)[/tex]
Step-by-step explanation:
[tex]\frac{(2n)!}{n!}[/tex] is interesting
if we write it out for n=4
[tex]\frac{(2n)!}{n!}=\frac{(2(4))!}{4!}=\frac{8!}{4!}=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5 \cdot 6\cdot 7\cdot 8}{1\cdot 2\cdot 3\cdot 4}=5\cdot 6\cdot 7\cdot 8[/tex]
try for n=3
[tex]\frac{(2n)!}{n!}=\frac{(2(3))!}{3!}=\frac{6!}{3!}=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5 \cdot 6}{1\cdot 2\cdot 3}=4\cdot 5\cdot 6[/tex]
we notice that the first [tex]n[/tex] terms cancel out
therefore, [tex]\frac{(2n)!}{n!}=(n+1)\cdot (n+2)\dots (2n)[/tex]
there is no further information given
