If $C_0, C_1, C_2, .., C_n$ are the binomial coefficients in the expansion of $(1+x)^n$ If $C_0, C_1, C_2,...,C_n$ are the binomial coefficients in the expansion of $(1+x)^n$, prove that:
$$C_{r}.C_{n} + C_{r+1}.C_{n-1} +......+ C_{n}.C_{r} = C(2n, n+r) =\dfrac {(2n)!}{(n-r)! (n+r)!}$$
Is there any way to approach this sort of questions using calculus (derivatives or integration)?
 A: THis suffices to differentiate $(1+x)^{2n}=(1+x)^n(1+x)^n $ $n+r$ times.
Differentiating the right side first, 
By (General Leibniz rule), it's derivative is given by differentiating the first part $k$ times and the second part $(n+r)-k$ times.
It's derivative is given by $\sum_{k=0}^{n+r} {{n}\choose{k}}((1+x)^n)^{(k)}((1+x)^n)^{(n+r-k)}$
$=\sum_{k=0}^{n-r} {{n}\choose{k+r}} ((1+x)^n)^{(r+k)}((1+x)^n)^{(n-k)}$
Evaluating at $x=0$,
$=\sum_{k=0}^{n-r} \dfrac{(n+r)!}{(k+r)!(n-k)!} \dfrac{n!}{(n-r-k)!}\dfrac{n!}{k!}$
$=(n+r)! \sum_{k=0}^{n-r} \dfrac{n!}{(k+r)!(n-r-k)!} \dfrac{n!}{(n-k)!k!}$
$=(n+r)! \sum_{k=0}^{n-r} {{n}\choose{k+r}}{{n}\choose{n-k}}$
Differentiating left side gives $(n+r)!{{2n}\choose{n+r}}$
Giving the desired equality.
A: We can solve this problem using algebra without needs for differentiation or integration. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
\begin{align*}
\binom{n}{k}=[x^k](1+x)^n\tag{1}
\end{align*}

We start with the right-hand side and obtain
  \begin{align*}
\color{blue}{\frac{(2n)!}{(n-r)!(n+r)!}}&=\binom{2n}{n+r}\\
&=[x^{n+r}](1+x)^{2n}\tag{2}\\
&=[x^{n+r}]\sum_{k=0}^n\binom{n}{k}x^k\sum_{l=0}^n\binom{n}{l}x^l\tag{3}\\
&=\sum_{k=0}^n\binom{n}{k}[x^{n+r-k}]\sum_{l=0}^n\binom{n}{l}x^l\tag{4}\\
&=\sum_{k=0}^n\binom{n}{k}\binom{n}{n+r-k}\tag{5}\\
&\,\,\color{blue}{=\sum_{k=r}^n\binom{n}{k}\binom{n}{n+r-k}}\tag{6}
\end{align*}
and the claim follows.

Comment:


*

*In (2) we write $\binom{2n}{n+r}$ using the coefficient of operator according to (1).

*In (3) we write $(1+x)^{2n}=(1+x)^n(1+x)^n$ and expand.

*In (4) we use the rule $[x^{p+q}]A(x)=[x^p]x^{-q}A(x)$.

*In (5) we select the coefficient of $x^{n+r-k}$.

*In (6) we note that $\binom{p}{q}=0$ if $q>p$ and set the lower index consequently to $k=r$.
