What is the sum of the given series? How do I sum the following series...
$$ \sum_{r = 0}^{40}{r\binom{40}{r}\binom{30}{r}}$$
When I expanded, I noticed, it turns out to be...
$$0\times\binom{40}{0}\binom{30}{0}+1\times\binom{40}{1}\binom{30}{1}+2\times\binom{40}{2}\binom{30}{2}+3\times\binom{40}{3}\binom{30}{3}+....$$and so on.
So I think that the approach can be to differentiate $$(x+1)^n $$ and its expansion so that the powers of the $x,x^2,x^3,...$ terms get multiplied in front as $1,2,3,4....$, but I can't figure out the approach for two different combinations multiplied together.
Also, for $r>30$, the $\binom{30}{r}$ terms become undefined right? So how do I handle those?
 A: For $r>0,$ $$r\binom{40}r=40\binom{39}{r-1}$$
Now consider $$(1+x)^{39}\left(1+\dfrac1x\right)^{30}=\dfrac{(1+x)^{69}}{x^{30}}$$
Compare the coefficients of $x^{39-(r-1)-(30-r)}$
A: $$
\eqalign{
  & \sum\limits_{r = 0}^{40} {r\left( \matrix{
  40 \cr 
  r \cr}  \right)\left( \matrix{
  30 \cr 
  r \cr}  \right)}  =   \cr 
  &  = \sum\limits_{r = 1}^{\left( {30} \right)} {r\left( \matrix{
  40 \cr 
  r \cr}  \right)\left( \matrix{
  30 \cr 
  r \cr}  \right)}  = \quad \quad (1)  \cr 
  &  = 40\sum\limits_{r = 1}^{30} {\left( \matrix{
  39 \cr 
  r - 1 \cr}  \right)\left( \matrix{
  30 \cr 
  r \cr}  \right)}  = \quad \quad (2)  \cr 
  &  = 40\sum\limits_{r = 0}^{29} {\left( \matrix{
  39 \cr 
  r \cr}  \right)\left( \matrix{
  30 \cr 
  r + 1 \cr}  \right)}  = \quad \quad (3)  \cr 
  &  = 40\sum\limits_{\left( {0\, \le } \right)\,r\,\left( { \le \,29} \right)} {\left( \matrix{
  39 \cr 
  r \cr}  \right)\left( \matrix{
  30 \cr 
  r + 1 \cr}  \right)}  = \quad \quad (4)  \cr 
  &  = 40\sum\limits_{\left( {0\, \le } \right)\,r\,\left( { \le \,29} \right)} {\left( \matrix{
  39 \cr 
  r \cr}  \right)\left( \matrix{
  30 \cr 
  29 - r \cr}  \right)}  = \quad \quad (5)  \cr 
  &  = 40\left( \matrix{
  69 \cr 
  29 \cr}  \right)\quad \quad (6) \cr} 
$$
where:
 (1) the lower actual index is 1 , since for $r=0$ the product is null,
and the upper is actually $30$ for the second binomial not to be null;
 (2) absorption;
 (3) change index;
 (4) the bounds are implicit in the binomials;
 (5) $0 \le 30$ and we can apply symmetry;
 (6) Vandermonde convolution
