Show $\sum_{k=1}^n {k+1\choose 2}{2n+1\choose n+k+1}={n\choose 1}4^{n-1}$ 
I've been attempting to show that:
  $$\sum_{k=1}^n {k+1\choose 2}{2n+1\choose n+k+1}={n\choose 1}4^{n-1}\\
\sum_{k=2}^n {k+2\choose 4}{2n+1\choose n+k+1}={n\choose 2}4^{n-2}$$

Can anyone give some direction?
 A: Let's rearrange the sum as
$$ 
\eqalign{ 
  & s(n) = \sum\limits_{k = 1}^n {\left( \matrix{ 
  k + 1 \cr  
  2 \cr}  \right)\left( \matrix{ 
  2n + 1 \cr  
  n + k + 1 \cr}  \right)}  =   \cr  
  &  = \sum\limits_{k = \,0}^{n - 1} {\left( \matrix{ 
  k + 2 \cr  
  2 \cr}  \right)\left( \matrix{ 
  2n + 1 \cr  
  n + k + 2 \cr}  \right)}  =   \cr  
  &  = \sum\limits_{0\, \le \,k\,} {\left( \matrix{ 
  k + 2 \cr  
  2 \cr}  \right)\left( \matrix{ 
  2n + 1 \cr  
  n + k + 2 \cr}  \right)}  \cr}  
$$ 
The summand is equal to 
$$ 
 \eqalign{ 
  & t_{\,k}  = \left( \matrix{  k + 2 \cr   2 \cr}  \right) 
 \left( \matrix{  2n + 1 \cr   n + k + 2 \cr}  \right) =   \cr  
  &  = {{\Gamma \left( {k + 3} \right)} \over {\Gamma \left( {k + 1} \right)\Gamma \left( {n + k + 3} \right)\Gamma \left( {n - k} \right)}} 
{{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( 3 \right)}} \cr}  
$$
so that 
$$ 
t_{\,0}  = \left( \matrix{ 
  2n + 1 \cr  
  n + 2 \cr}  \right) = {{\Gamma \left( {2n + 2} \right)}  
 \over {\Gamma \left( {n + 3} \right)\Gamma \left( n \right)}} 
$$
and the ratio is
$$ 
\eqalign{ 
  & {{t_{\,k + 1} } \over {t_{\,k} }} =   \cr  
  &  = {{\Gamma \left( {k + 4} \right)} 
 \over {\Gamma \left( {k + 2} \right)\Gamma \left( {n + k + 4} \right)\Gamma \left( {n - 1 - k} \right)}} 
 {{\Gamma \left( {k + 1} \right)\Gamma \left( {n + k + 3} \right)\Gamma \left( {n - k} \right)} 
 \over {\Gamma \left( {k + 3} \right)}} =   \cr  
  &  = {{\Gamma \left( {k + 4} \right)\Gamma \left( {k + 1} \right)\Gamma \left( {n + k + 3} \right)\Gamma \left( {n - k} \right)} 
 \over {\Gamma \left( {k + 3} \right)\Gamma \left( {k + 2} \right)\Gamma \left( {n + k + 4} \right)\Gamma \left( {n - 1 - k} \right)}} =   \cr  
  &  = {{\left( {k + 3} \right)\left( {n - 1 - k} \right)} \over {\left( {k + 1} \right)\left( {n + k + 3} \right)}} 
 =  - {{\left( {k + 3} \right)\left( {k - \left( {n - 1} \right)} \right)} \over {\left( {k + 1} \right)\left( {k + n + 3} \right)}} \cr}  
$$ 
That means that we can write $s(n))$ in terms of a Hypergeometric function
$$ 
s(n) = {{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( {n + 3} \right)\Gamma \left( n \right)}} 
{}_2F_{\,1} \left( {\left. {\matrix{ 
   {\;3,\; - \left( {n - 1} \right)}  \cr  
   {n + 3}  \cr  
 } \;} \right|\; - 1} \right) 
$$ 
Since
$$ 
1 + a - b = 1 + 3 + n - 1 = n + 3 = c 
$$
we can apply [Kummer's theorem](https://en.wikipedia.org/wiki/Hypergeometric_function#Kummer's_theorem_(z_=_%E2%88%921) 
and get 
$$ 
\eqalign{ 
  & s(n) = {{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( {n + 3} \right)\Gamma \left( n \right)}}{}_2F_{\,1} \left( {\left. {\matrix{ 
   {\;3,\; - \left( {n - 1} \right)}  \cr  
   {n + 3}  \cr  
 } \;} \right|\; - 1} \right) =   \cr  
  &  = {{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( {n + 3} \right)\Gamma \left( n \right)}}{{\Gamma \left( {n + 3} \right)\Gamma \left( {5/2} \right)} 
 \over {\Gamma \left( 4 \right)\Gamma \left( {3/2 + n} \right)}} =   \cr  
  &  = {{\Gamma \left( {5/2} \right)} \over {\Gamma \left( 4 \right)}}{{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( n \right)\Gamma \left( {3/2 + n} \right)}} \cr}  
$$ 
Then by the duplication formula 
$$ 
\Gamma \left( {2\,n + 2} \right) = 2^{\,2\,n + 1} {{\Gamma \left( {n + 1} \right)\Gamma \left( {n + 3/2} \right)} \over {\Gamma \left( {1/2} \right)}} 
$$
we finally reach to 
$$ 
\eqalign{ 
  & s(n) = {{\Gamma \left( {5/2} \right)} \over {\Gamma \left( 4 \right)\Gamma \left( {1/2} \right)}}{{\Gamma \left( {n + 1} \right)\Gamma \left( {n + 3/2} \right)} 
 \over {\Gamma \left( n \right)\Gamma \left( {3/2 + n} \right)}}2^{\,2\,n + 1}  =   \cr  
  &  = {{\Gamma \left( {5/2} \right)} \over {\Gamma \left( 4 \right)\Gamma \left( {1/2} \right)}}n\,2^{\,2\,n + 1}  = {1 \over 8}n\,2^{\,2\,n + 1}  = n\,2^{\,2\,n - 2}  \cr}  
$$ 
