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the third line where n=2. How did we determine that it can be written as...
\[\frac{C_0}{1\cdot2\cdot3\cdot4}\]

Is that Taylor series or something? D:

no not really the original function is
\[C_{n+2}=-\frac{C_n}{(n+2)(n+1)}\]

I'm sure @UnkleRhaukus can help...where is he?

yeah I think you're right, Thanks!

It is straight forward I know...LOL and Turing has taught me how to do this...I just forgot

I guess doing math and watching The Walking Dead can prevent one from doing simple math ;P

@UnkleRhaukus oooops, yeah good call :)