<s>[INST] Summation of binomial coefficients upon $(r+2)$ 
$$\sum^{50}_{r=0}(-1)^r \dfrac{\dbinom {50}r}{r+2}= ?$$

Attempt: 
$(1-x)^{50}= \sum (-1)^r \dbinom{50}r x^{n-r}$
Integrating both sides and then placing limits 0 to 1:
$\dfrac{1}{51}= \displaystyle \sum_{r=0}^{50}\dfrac{(-1)^r \dbinom{50}r}{52-r}$
So the answer should be $\dfrac{1}{51}$
But answer given is $1/(51\times 52)$
Where have I gone wrong?
Edit: 
I understood my mistake from the comment. 
Now I have:
$\dfrac{1}{51}= \displaystyle \sum_{r=0}^{50}\dfrac{(-1)^r \dbinom{50}r}{51-r}$
Is it possible to complete it from here? 
 [/INST] 
OP's expression is a special case of a formula for the reciprocal binomial coefficient using the Beta function
\begin{align*}
\color{blue}{\binom{n}{k}^{-1}}&\color{blue}{=(n+1)\int_0^1 z^k(1-z)^{n-k}\,dz}\\
&=(n+1)\int_0^1z^k\sum_{r=0}^{n-k}\binom{n-k}{r}(-z)^r\,dz\\
&=(n+1)\sum_{r=0}^{n-k}\binom{n-k}{r}(-1)^r\int_0^1z^{k+r}\,dz\\
&\,\,\color{blue}{=(n+1)\sum_{r=0}^{n-k}\binom{n-k}{r}\frac{(-1)^r}{k+r+1}}\tag{1}
\end{align*}

We obtain from (1)
\begin{align*}
\sum_{r=0}^{n-k}\binom{n-k}{r}\frac{(-1)^r}{k+r+1}=\frac{1}{(n+1)\binom{n}{k}}\tag{2}
\end{align*}

Putting $n=51$ and $k=1$ in (2) we derive OP's formula
  \begin{align*}
\color{blue}{\sum_{r=0}^{50}\binom{50}{r}\frac{(-1)^r}{r+2}}=\frac{1}{52\binom{51}{1}}\color{blue}{=\frac{1}{52\cdot 51}}
\end{align*}

</s>