Q:

Closed form for a combinatorial series - Convolution.

Is there any closed form for this series?
$$t+(t-1)\times {d \choose 1}+(t-2)\times {d \choose 2}+ \cdots +2\times {d \choose t-2}+ {d \choose t-1},$$
where $t<\frac d2$.
I think there are some useful techniques of Riordan Array but I don't know exactly what to do.

A:

We have
$$
\eqalign{
  & f(t,d) = \sum\limits_{0\, \le \,k\, \le \,t - 1} {\left( {t - k} \right)\left( \matrix{
  d \cr 
  k \cr}  \right)}  = \sum\limits_{0\, \le \,k\, \le \,t} {\left( {t - k} \right)\left( \matrix{
  d \cr 
  k \cr}  \right)}  =   \cr 
  &  = t\sum\limits_{0\, \le \,k\, \le \,t} {\left( \matrix{
  d \cr 
  k \cr}  \right)}  - \sum\limits_{0\, \le \,k\, \le \,t} {k\left( \matrix{
  d \cr 
  k \cr}  \right)}  = t\sum\limits_{0\, \le \,k\, \le \,t} {\left( \matrix{
  d \cr 
  k \cr}  \right)}  - d\sum\limits_{0\, \le \,k\, \le \,t} {\left( \matrix{
  d - 1 \cr 
  k - 1 \cr}  \right)}  =   \cr 
  &  = t\sum\limits_{0\, \le \,k\, \le \,t} {\left( \matrix{
  d \cr 
  k \cr}  \right)}  - d\sum\limits_{0\, \le \,k\, \le \,t - 1} {\left( \matrix{
  d - 1 \cr 
  k \cr}  \right)}  \cr} 
$$
Unfortunately there is no closed form for the partial sum of Binomials 
done on the lower index, unless invoking Hypergeometric functions or generating functions.  
Concerning the generating function, as you rightly noted we have a convolution, between
the sequences $\{0,1, \cdots, n, \cdots \}$ and $\{ \binom{d,0},\binom{d,1},\cdots ,\binom{d,n},\cdots \}$.
Since
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {n\,z^{\,n} }  = z\sum\limits_{0\, \le \,n} {n\,z^{\,n - 1} }
  = z{d \over {dz}}\sum\limits_{0\, \le \,n} {\,z^{\,n} }  = {z \over {\left( {1 - z} \right)^{\,2} }}  \cr 
  & \sum\limits_{0\, \le \,n} {\left( \matrix{
  d \cr 
  n \cr}  \right)z^{\,n} }  = \left( {1 + z} \right)^{\,d}  \cr} 
$$
then
$$
F(z) = \sum\limits_{0\, \le \,t} {\left( {\sum\limits_{0\, \le \,k\, \le \,t} {\left( {t - k} \right)\left( \matrix{
  d \cr 
  k \cr}  \right)} } \right)z^{\,t} }  = {{z\left( {1 + z} \right)^{\,d} } \over {\left( {1 - z} \right)^{\,2} }}
$$
Concerning instead the expression through the hypergeometric function, we have
$$
\eqalign{
  & \sum\limits_{0\, \le \,k\, \le \,t} {\left( \matrix{
  d \cr 
  k \cr}  \right)}  = 2^{\,d}  - \sum\limits_{t + 1\, \le \,k} {\left( \matrix{
  d \cr 
  k \cr}  \right)}  = 2^{\,d}  - \sum\limits_{0\, \le \,k} {\left( \matrix{
  d \cr 
  k + t + 1 \cr}  \right)}  =   \cr 
  &  = 2^{\,d}  - \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k + t + 1} \left( \matrix{
  k + t - d \cr 
  k + t + 1 \cr}  \right)}  = 2^{\,d}  - \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k + t + 1} {{\left( {k + t - d} \right)^{\,\underline {\,k + t + 1\,} } } \over {\left( {k + t + 1} \right)!}}}  =   \cr 
  &  = 2^{\,d}  - \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k + t + 1} {{\left( {k + t - d} \right)^{\,\underline {\,k\,} } \left( {t - d} \right)^{\,\underline {\,t + 1\,} } } \over {1^{\,\overline {\,k + t + 1\,} } }}}  =   \cr 
  &  = 2^{\,d}  - \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k + t + 1} {{\left( {t - d + 1} \right)^{\,\overline {\,k\,} } \left( {t - d} \right)^{\,\underline {\,t + 1\,} } } \over {1^{\,\overline {\,t + 1\,} } \left( {t + 2} \right)^{\,\overline {\,k\,} } }}}  =   \cr 
  &  = 2^{\,d}  - \left( { - 1} \right)^{t + 1} \left( \matrix{
  t - d \cr 
  t + 1 \cr}  \right)\sum\limits_{0\, \le \,k} {{{1^{\,\overline {\,k\,} } \left( {t - d + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {t + 2} \right)^{\,\overline {\,k\,} } }}{{\left( { - 1} \right)^{\,k} } \over {k!}}}  =   \cr 
  &  = 2^{\,d}  - \left( \matrix{
  d \cr 
  t + 1 \cr}  \right)\sum\limits_{0\, \le \,k} {{{1^{\,\overline {\,k\,} } \left( {t - d + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {t + 2} \right)^{\,\overline {\,k\,} } }}{{\left( { - 1} \right)^{\,k} } \over {k!}}}  =   \cr 
  &  = 2^{\,d}  - \left( \matrix{
  d \cr 
  t + 1 \cr}  \right){}_2F_1 (1,t - d + 1;\;t + 2;\; - 1) \cr} 
$$
where the overlined and underlined exponents indicate respectively the Rising and the Falling Factorial.
Then using the result of the first identity above we can easily express
your sum by means of combination of two hypergeometrics. 
Finally, in order to answer to your question about the minimum $t$ such that 
$2^d \le f(t,d)$ we shall try and  express $f(t,d)$ as an explicit function of $t$.      
Now the binomial $\binom{d}{k}$, expressed as a continuous function in terms of Gamma,
has a bell-shaped graph, symmetric around $d/2$, which resembles a Gaussian.
In fact for large positive $d$ we can express it as
$$
\eqalign{
  & \binom{d}{d/2+x} = \binom{d}{d/2-x} = \cr
  &  = \binom{d}{d/2}
{{\Gamma (1 + d/2)^{\,2} } \over {\Gamma (1 + d/2 + x)\Gamma (1 + d/2 - x)}}\quad \buildrel {d\, \to \,\infty } \over
 \longrightarrow   \cr 
  &  \approx \binom{d}{d/2}\exp \left( { - \psi ^{\,\left( 1 \right)} (1 + d/2)\;x^{\,2} } \right)\;
 \approx \;\binom{d}{d/2} \exp \left( { - 2\;x^{\,2} /d} \right)\; \approx   \cr 
  &  \approx 2^{\,d} {1 \over {\sqrt {2\pi \,\left( {\sqrt d /2} \right)^{\,2} } }}\exp \left( { - \;x^{\,2} /2\left( {\sqrt d /2} \right)^{\,2} } \right) = 2^{\,d} {\cal N}(0,d/4) \cr} 
$$
Now, the sum of the binomial
$$
\sum\limits_{0\, \le \,k\, \le \,t} {\binom{d}{k}}
$$
is $0$ at $t=-1$, $1$ at $t=0$, and finishing at $2^d$ at $t=d$. That means that the median
happens to be at $(d-1)/2$ instead than at $d/2$.
Therefore we can approximate it with the integral of the Normal shifted to have $\mu=(d-1)/2$, i.e.:
$$
{1 \over {2^{\,d} }}\sum\limits_{0\, \le \,k\, \le \,t} {\binom{d}{k} }
 \approx \int_{ - \infty }^{\,t} {{1 \over {\sqrt {2\pi \,\left( {\sqrt d /2} \right)^{\,2} } }}
 \exp \left( { - \;\left( {x - \left( {d - 1} \right)/2} \right)^{\,2} \mathop /\limits_{} \left( {2\left( {\sqrt d /2} \right)^{\,2} } \right)} \right)dx} 
$$
It is difficult to assess  analytically the goodness of such an approximation, however the simple sketch 
below can give some indications ( the points represent the sum and the lines  the integral).

Then, of course, it depends on what you have actually need to attain.

