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I did this problem a month ago and I somewhat forgot how to do it.

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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 \[C_{n+2}=-\frac{C_n}{(n+2)(n+1)}\]

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Sorry. This goes over my dumb head. :'( Math gods like @satellite73 @amistre64 @asnaseer can help you with problems like these. :/
I'm sure @UnkleRhaukus can help...where is he?
@eseidl @eSpeX can you guys solve this? Last time I saw Rhau in chem section.
Am I missing something? Isn't it just:\[C_{n+2}=-\frac{c_n}{(n+2)(n+1)}\]Input n=2\[C_{2+2}=-\frac{c_2}{(2+2)(2+1)}\]So,\[C_4=-\frac{c_2}{4*3}\]But we know\[c_2=-\frac{c_0}{1*2}\]From the table where n=0.Thus:\[C_4=-\frac{c_0}{4*3*2*1}=-\frac{c_0}{4!}\]
yeah I think you're right, Thanks!
lol, yeah it just seemed too straight forward to call in the "math gods" (which I definitely am not one of lol)
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
@eseidl all you missing is the negative signs cancel \[n=2\qquad C_4=\frac{-C_2}{4\cdot3}=-\frac{-\tfrac{C_0}{2\cdot1}}{4\cdot3}=\frac{C_0}{4\cdot3\cdot 2\cdot1}=\frac{C_0}{4!}\],
@UnkleRhaukus oooops, yeah good call :)

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