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JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0the third line where n=2. How did we determine that it can be written as... \[\frac{C_0}{1\cdot2\cdot3\cdot4}\]

saifoo.khan
 2 years ago
Best ResponseYou've already chosen the best response.0Is that Taylor series or something? D:

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0no not really the original function is \[C_{n+2}=\frac{C_n}{(n+2)(n+1)}\]

saifoo.khan
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry. This goes over my dumb head. :'( Math gods like @satellite73 @amistre64 @asnaseer can help you with problems like these. :/

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I'm sure @UnkleRhaukus can help...where is he?

saifoo.khan
 2 years ago
Best ResponseYou've already chosen the best response.0@eseidl @eSpeX can you guys solve this? Last time I saw Rhau in chem section.

eseidl
 2 years ago
Best ResponseYou've already chosen the best response.2Am 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:\[C_4=\frac{c_0}{4*3*2*1}=\frac{c_0}{4!}\]

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0yeah I think you're right, Thanks!

eseidl
 2 years ago
Best ResponseYou've already chosen the best response.2lol, yeah it just seemed too straight forward to call in the "math gods" (which I definitely am not one of lol)

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0It is straight forward I know...LOL and Turing has taught me how to do this...I just forgot

JenniferSmart1
 2 years ago
Best ResponseYou've already chosen the best response.0I guess doing math and watching The Walking Dead can prevent one from doing simple math ;P

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1@eseidl all you missing is the negative signs cancel \[n=2\qquad C_4=\frac{C_2}{4\cdot3}=\frac{\tfrac{C_0}{2\cdot1}}{4\cdot3}=\frac{C_0}{4\cdot3\cdot 2\cdot1}=\frac{C_0}{4!}\],

eseidl
 2 years ago
Best ResponseYou've already chosen the best response.2@UnkleRhaukus oooops, yeah good call :)
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