## JenniferSmart1 Group Title I did this problem a month ago and I somewhat forgot how to do it. one year ago one year ago

1. JenniferSmart1 Group Title

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

2. saifoo.khan Group Title

Is that Taylor series or something? D:

3. JenniferSmart1 Group Title

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

4. saifoo.khan Group Title

Sorry. This goes over my dumb head. :'( Math gods like @satellite73 @amistre64 @asnaseer can help you with problems like these. :/

5. JenniferSmart1 Group Title

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

6. saifoo.khan Group Title

@eseidl @eSpeX can you guys solve this? Last time I saw Rhau in chem section.

7. eseidl Group Title

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:$C_4=-\frac{c_0}{4*3*2*1}=-\frac{c_0}{4!}$

8. JenniferSmart1 Group Title

yeah I think you're right, Thanks!

9. eseidl Group Title

lol, yeah it just seemed too straight forward to call in the "math gods" (which I definitely am not one of lol)

10. JenniferSmart1 Group Title

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

11. JenniferSmart1 Group Title

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

12. UnkleRhaukus Group Title

@eseidl all you missing is the negative signs cancel $n=2\qquad C_4=\frac{-C_2}{4\cdot3}=-\frac{-\tfrac{C_0}{2\cdot1}}{4\cdot3}=\frac{C_0}{4\cdot3\cdot 2\cdot1}=\frac{C_0}{4!}$,

13. eseidl Group Title

@UnkleRhaukus oooops, yeah good call :)