Find the value of $ \sum _{r=0} ^{2n} r ( ^{2n}C _r) ( \frac 1{r+2} ) $ 
Find the value of $$ \sum _{r=0} ^{2n} r ( ^{2n}C _r ) ( \frac 1{r+2} )$$


In order to solve this I am trying to make the term(s) of the series independent of $r$. However I'm unable to solve this further: $$ \sum _{r=0} ^{2n} 2n (^{2n-1}C_{r-1})( \frac 1{r+2} ) $$ 
Any help would be appreciated. :)
 A: HINT:
$$r\binom{2n}r\cdot\frac1{r+2}=\frac{r(r+1)}{(2n+2)(2n+1)}\cdot\frac{(2n+2)!}{[(2n+2)-(r+2)]!\cdot(r+2)!}$$
$$=\frac{r(r+1)}{(2n+2)(2n+1)}\cdot\binom{2n+2}{r+2}$$
Now let $r(r+1)=(r+2)(r+1)+A(r+2)+B$
$r^2+r=r^2+r(3+A)+2+2A+B$
$\implies A+3=0\iff A=-3$
and $B+2A+2=0\iff B=-2A-2=4$
$$\implies(2n+2)(2n+1)r\cdot \binom{2n}r\cdot\frac1{r+2}=[(r+2)(r+1)+(-3)(r+2)+4]\binom{2n+2}{r+2}$$
$$=(r+2)(r+1)\binom{2n+2}{r+2}-3(r+2)\binom{2n+2}{r+2}+4\binom{2n+2}{r+2}$$
$$=(2n+2)(2n+1)\binom{2n}r-3(2n+1)\binom{2n+1}{r+1}+4\binom{2n+2}{r+2}$$
Finally, $\sum_{r=0}^{2n}\binom{2n}r=(1+1)^{2n}$
and  $\sum_{r=0}^{2n}\binom{2n+1}{r+1}=(1+1)^{2n+1}-\binom{2n+1}0$
and  $\sum_{r=0}^{2n}\binom{2n+2}{r+2}=(1+1)^{2n+2}-\binom{2n+2}0-\binom{2n+2}1$
A: Do you mean $\sum_{r=0}^{2n} \dfrac{r}{r+2} {2n \choose r}$?  Maple says it's
$${\frac {2\,{4}^{n}{n}^{2}-{4}^{n}n+{4}^{n}-1}{ \left( 2\,n+1 \right) 
 \left( n+1 \right) }}
$$
A: Here is an approach. Differentiating the identity

$$ S = \sum _{r=0} ^{2n} ( ^{2n}C _r) x^{r} = (1+x)^{2n}. $$

with respect to $x$ gives

$$ S'(x) = \sum _{r=0} ^{2n} r( ^{2n}C _r) x^{r-1}= 2n ( 1+x )^{2n-1} \longrightarrow (1).$$

Multiplying $(1)$ by $x^2$, integrating from $0$ to $1$ w.r.t. $x$ and then substituting $x=1$ gives the desired result

$$ \sum _{r=0} ^{2n} r( ^{2n}C _r) \frac{x^{r+2}}{r+2} = 2n\int_{0}^{1} x^2(1+x)^{2n-1} dx=\dots\,. $$

I let you do the integration and finish the problem.
