$\begingroup$
$\endgroup$
2
Can you explain me how we got this identity?
$$\frac{1}{(3n)!}$$ the same as $$\frac{(3n)!}{(3n+3)!}$$ I have been trying to expand, but didn't get the same.
Thanks.
2 Answers
$\begingroup$
$\endgroup$
1
they are not same , consider $n=1$ , you have $\frac{1}{3!}=\frac{1}{6}$ and on the other hand $\frac{3!}{6!}= \frac{1}{120}$
$\begingroup$
$\endgroup$
1
If you expand $$(3n+3)!=(3n+3)\times(3n+2)\times(3n+1)\times(3n)!$$ So $$\frac{(3n)!}{(3n+3)!}=\frac 1 {(3n+3)\times(3n+2)\times(3n+1)}$$
