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The sum of the series

from r=1 to r=3n-1
where n is an even natural numeber

Killing question don't post these no one can solve this..


 \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)!}
=\frac{3n+1}{(3n+2)} (\frac{(3n+1-r)!r!}{(3n+1)!}+ \frac{(3n-r)!(r+1)!}{(3n+1)!} )

= \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.

2 5 2
very complex!!
wait i will complete it
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