Sum of multiplication of binomial with consecutive rising term I am looking for a closed form for the following sum:
$$\sum_{p=0}^{k} \left( \begin{matrix} n \\ p+1 \end{matrix} \right) \left( \begin{matrix} k \\ p \end{matrix} \right) 4^p = \;\;???$$
I have tried to expand:
$$(1+2x)^n \cdot \left( 2+\frac{1}{x}\right)^k$$
without success. Any ideas are welcome!
Thank you
 A: Please allow me to reserve the use of $k$ as an index and rewrite your sum as
$$
\eqalign{
  &S =  \sum\limits_{k = 0}^m {\left( \matrix{
  n \cr 
  k + 1 \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  = \sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {\left( \matrix{
  n \cr 
  k + 1 \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {\left( \matrix{
  n \cr 
  n - k - 1 \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  \cr} 
$$
A first approach is to consider that the double sum of the binomial correlation is
$$
\eqalign{
  & S = \sum\limits_{k = 0}^m {\left( \matrix{
  n \cr 
  k + 1 \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  = \sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {\left( \matrix{
  n \cr 
  k + 1 \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {\left( \matrix{
  n \cr 
  n - k - 1 \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  \cr} 
$$
So that S is the coefficient of $x^{n-1}$ in the binomial
$$
S = \left[ {x^{n - 1} } \right]\left( {\left( {1 + 4x} \right)^{\,m} \left( {1 + x} \right)^{\,n} } \right)
$$
We can also express the sum directly in terms of the Hypergeometric function as
$$
\eqalign{
  & S = \sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {\left( \matrix{
  n \cr 
  k + 1 \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  =   \cr 
  &  = n\sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {{1 \over {k + 1}}\left( \matrix{
  n - 1 \cr 
  k \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  =   \cr 
  &  = n\sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {{1 \over {k + 1}}\left( \matrix{
  n - 1 \cr 
  k \cr}  \right)\left( \matrix{
  m \cr 
  k \cr}  \right)4^{\,k} }  = n\sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {{{\left( {n - 1} \right)^{\,\underline {\,k\,} } m^{\,\underline {\,k\,} } } \over {\left( {k + 1} \right)!k!}}4^{\,k} }  =   \cr 
  &  = n\sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {{{\left( { - 1} \right)^{\,k} \left( { - n + 1} \right)^{\,\overline {\,k\,} } \left( { - 1} \right)^{\,k} \left( { - m} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k + 1\,} } 1^{\,\overline {\,k\,} } }}4^{\,k} }  =   \cr 
  &  = n\sum\limits_{0\, \le \,k\,\left( { \le \,m,n - 1} \right)} {{{\left( { - n + 1} \right)^{\,\overline {\,k\,} } \left( { - m} \right)^{\,\overline {\,k\,} } } \over {2^{\,\overline {\,k\,} } }}{{4^{\,k} } \over {1^{\,\overline {\,k\,} } }}}  =   \cr 
  &  = n\;{}_2F_{\,1} \left( {\left. {\matrix{
   { - n + 1, - m}  \cr 
   2  \cr 
 } \;} \right|\;4} \right) \cr} 
$$
