Unusual binomial sum: $\sum_{d=k}^{n} {d \choose k} p^{d}(1-p)^{n-d}$ Does anyone know how to simplify the following sum? It's been giving me and everyone else I've showed it to quite a bit of trouble. I'm quite confident that this should simplify, but I just can't seem to see how. \begin{equation} \sum_{d=k}^{n} {d \choose k} p^{d}(1-p)^{n-d}
\end{equation}
Where k is greater than 1. Please note that the sum varies over the upper index of the binomial. The attempts thus far have pretty much all involved expressing the terms as a series and trying to relate that series to the product of the derivatives of order up to and including d. This method is proving intractable as it only works for k=2, otherwise we get the product of the product (k-1 times) of a series involving higher order derivatives, which is hardly a simplification.
Any and all insights are welcome. 

We can compute the generating function
$$
\begin{align}
\sum_{n=k}^\infty\sum_{d=k}^n\binom{d}{k}p^d(1-p)^{n-d}x^n
&=\sum_{d=k}^\infty\sum_{n=d}^\infty\binom{d}{k}p^d(1-p)^{n-d}x^n\\
&=\sum_{d=k}^\infty\binom{d}{k}\frac{p^dx^d}{1-(1-p)x}\\
&=(px)^k\sum_{d=0}^\infty\binom{-k-1}{d}\frac{(-1)^dp^dx^d}{1-(1-p)x}\\
&=\frac{(px)^k}{(1-px)^{k+1}}\frac1{1-(1-p)x}\tag{1}
\end{align}
$$
In the special case of $p=\frac12$, we get
$$
\begin{align}
\frac{\left(\frac12x\right)^k}{\left(1-\frac12x\right)^{k+2}}
&=\sum_{j=0}^\infty\binom{-k-2}{j}(-1)^j\left(\tfrac12x\right)^{k+j}\\
&=\sum_{j=0}^\infty\binom{k+j+1}{j}\left(\tfrac12x\right)^{k+j}\\
&=\sum_{n=k}^\infty\binom{n+1}{k+1}\left(\tfrac12x\right)^n\tag{2}
\end{align}
$$
From $(1)$, we get
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{d=k}^n\binom{d}{k}p^d(1-p)^{n-d}=\left[x^{n-k}\right]\left(\frac{p^k}{(1-px)^{k+1}}\frac1{1-(1-p)x}\right)}\tag{3}
$$
In the special case of $p=\frac12$, $(2)$ gives
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{d=k}^n\binom{d}{k}\left(\frac12\right)^d\left(\frac12\right)^{n-d}=\binom{n+1}{k+1}\left(\frac12\right)^n}\tag{4}
$$
