Question

the sum of the series 
$\frac{(-1)^{r-1}}{C^{3n}_{r}}$
 from r=1 to r=3n-1
where n is an even natural numeber

Answer 1

take 
$\frac{1}{C^{3n}_{r}}= \frac{(3n-r)!r!}{3n!}= \frac{3n+1}{3n+2} \frac{3n+2}{3n+1} \frac{(3n-r)!r!}{3n!}$
$=\frac{(3n-r)!r!}{3n!}= \frac{3n+1}{(3n+2)} \frac{(3n+1-r)+(r+1)}{3n+1} \frac{(3n-r)!r!}{3n!}$
$=\frac{3n+1}{(3n+2)} \frac{(3n+1-r)+(r+1)(3n-r)!r!}{(3n+1)!}$
[tex]=\frac{3n+1}{(3n+2)} (\frac{(3n+1-r)!r!}{(3n+1)!}+ \frac{(3n-r)!(r+1)!}{(3n+1)!} ) [/tex]
$= \frac{3n+1}{3n+2}( \frac{1}{C_{r}^{3n+1} }+\frac{1}{C_{r+1}^{3n} })$
so now ur sum can be given as
$=(-1)^rr \frac{3n+1}{3n+2}( \frac{1}{C_{r}^{3n+1} }+\frac{1}{C_{r+1}^{3n} })$
so now the sum will be as follows
$=\frac{3n+1}{3n+2}( \frac{1}{C_{1}^{3n+1} }+\frac{1}{C_{2}^{3n}}-2( \frac{1}{C_{2}^{3n+1} }+\frac{1}{C_{3}^{3n} })+...)$
i am able to till that only.