Find $\sum_{r=0}^{2n}\frac{r}{r+2}\binom{2n}r\;.$ Find 
$$\sum_{r=0}^{2n}\frac{r}{r+2}\binom{2n}r\;.$$
I got 
$$\frac{2^{2n+1}(2n^2+n+1)-1}{(2n+1)(2n+2)}$$
but the answer is 
$$\frac{2^{2n+1}(2n^2-n+1)-2}{(2n+1)(2n+2)}$$
Thanks for the help...
 A: Here is a proof. We start with the identity

$$ \sum_{r=0}^{2n} {2n \choose r} x^r = (1+x)^{2n}.$$

Differentiating both sides w.r.t. $x$ gives
$$ \sum_{r=0}^{2n} r{2n \choose r} x^{r-1} = 2n(1+x)^{2n-1}.$$
Multiplying both sides by $x^2$ yields 
$$ \sum_{r=0}^{2n} r{2n \choose r} x^{r+1} = 2n\,x^2(1+x)^{2n-1}.$$
Integrating both sides w.r.t $x$ from $0$ to $1$ gives the desired result

$$ \sum_{r=0}^{2n} \frac{r}{r+2}{2n \choose r} = {\frac {2\,{4}^{n}{n}^{2}-{4}^{n}n+{4}^{n}-1}{(2n+1)(2n+2)}}.  $$

Note: You can evaluate the integral by using integration by parts by assuming $u=x^2$.
A: Surprisingly, I obtained another result which differs from both
$$\frac{2^{2n+1}(2n^2-n+1)-2}{(2n+1)(2n+2)}$$
and I am just unable to find the mistake(s).
I did not simplify my expression in order to stay as close as possible to those given in the post.
A: 
Here is a full proof, using a probabilistic interpretation.

Let $S_n$ denote the $n$th sum and $T_n=2^{-2n-2}(2n+1)(2n+2)S_n$, then
$$
T_n=2^{-2n-2}\sum_{r=0}^{2n}r(r+1){2n+2\choose r+2}=\sum_{r=0}^{2n}r(r+1)p_n(r+2),
$$
where, for every $0\leqslant s\leqslant2n+2$,
$$
p_n(s)=2^{-2n-2}{2n+2\choose s}.
$$
Thus, $p_n$ is the binomial distribution of parameters $(2n+2,\frac12)$, and, using the change of variable $s=r+2$, one gets
$$
T_n=\sum_{s=2}^{2n+2}(s-1)(s-2)p_n(s)=\sum_{s=0}^{2n+2}(s-1)(s-2)p_n(s)-2p_n(0),
$$
that is,
$$
T_n=E[(X_n-1)(X_n-2)]-2^{-2n-1},
$$
where $X_n$ is a random variable with distribution $p_n$. 
Now, $E[X_n]=n+1$ and $\mathrm{var}(X_n)=\frac12(n+1)$ hence $E[X_n^2]=\mathrm{var}(X_n)+E[X_n]^2$ is $E[X_n^2]=\frac12(n+1)(2n+3)$ and
$$
E[(X_n-1)(X_n-2)]=E[X_n^2]-3E[X_n]+2=\frac12(2n^2-n+1).
$$
Finally,

$$
S_n=2^{2n+2}\frac{\frac12(2n^2-n+1)-2^{-2n-1}}{(2n+1)(2n+2)}=\frac{4^{n}(2n^2-n+1)-1}{(2n+1)(n+1)}.
$$

A: $$\frac r{r+2}\binom{2n}r=\left(1-\frac2{r+2}\right)\binom{2n}r=\binom{2n}r-2\cdot\underbrace{\frac1{r+2}\binom{2n}r}_1$$
$$\text{Now,   }\underbrace{\frac1{r+2}\binom{2n}r}_1=(r+1)\cdot\frac{(2n)!}{(2n-r)!\cdot(r+2)(r+1) \cdot r!}$$
$$=\frac{r+1}{(2n+2)(2n+1)}\cdot\frac{(2n+2)!}{\{2n+2-(r+2)\}!\cdot(r+2)!} =\underbrace{\frac{r+1}{(2n+2)(2n+1)}\binom{2n+2}{r+2}}_2$$
$$\text{Again, }\underbrace{\frac{r+1}{(2n+2)(2n+1)}\binom{2n+2}{r+2}}_2=\frac{r+2-1}{(2n+2)(2n+1)}\binom{2n+2}{r+2}$$
$$=\underbrace{\frac{r+2}{2n+2)(2n+1)}\binom{2n+2}{r+2}}_3-\frac{1}{(2n+2)(2n+1)}\binom{2n+2}{r+2}$$
by distributive law.
Now $\displaystyle\underbrace{(r+2)\binom{2n+2}{r+2}}_3=(r+2)\frac{(2n+2)!}{(2n-r)! (r+2)!}=(2n+2)\frac{(2n+1)!}{\{(2n+1)-(r+1)\}! (r+1)!}=(2n+2)\binom{2n+1}{r+1}$
$\displaystyle\implies\frac r{r+2}\binom{2n}r$
$\displaystyle=\binom{2n}r-\frac2{(2n+2)(2n+1)}\left[(2n+2)\cdot\binom{2n+1}{r+1}-\binom{2n+2}{r+2}\right]\  \ \  \ (4)$
Now using $\displaystyle2^m=(1+1)^m=\sum_{0\le r\le m}\binom mr,$
$\displaystyle(i)\sum_{0\le r\le 2n}\binom{2n}r=2^{2n}$
$\displaystyle(ii)\sum_{0\le r\le 2n}\binom{2n+1}{r+1}=\sum_{0\le u\le 2n+1}\binom{2n+1}u-\binom{2n+1}0=2^{2n+1}-1$
$\displaystyle(iii)\sum_{0\le r\le 2n}\binom{2n+2}{r+2}=\sum_{0\le v\le 2n+2}\binom{2n+2}v-\binom{2n+2}0-\binom{2n+2}1$
$\displaystyle=2^{2n+2}-1-(2n+2)$
Use $(i),(ii),(iii)$ in $(4)$
