Unusual binomial sum: $\sum_{d=k}^{n} {d \choose k} p^{d}(1-p)^{n-d}$

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 $$ \bbox[5px,border:2px solid #C0A000]{\sum_{d=k}^n\binom{d}{k}\left(\frac12\right)^d\left(\frac12\right)^{n-d}=\binom{n+1}{k+1}\left(\frac12\right)^n}\tag{4} $$

Note: Here's is an answer providing a closed formula for the (simple) special case $p=\frac{1}{2}$. Since the used approach is often helpful to also find a general solution, it indicates that there is presumably no closed formula of OPs expression $$\sum_{d=k}^{n}\binom{d}{k}p^d(1-p)^{n-d}\qquad\qquad 0\leq k\leq n$$

We show the following is valid for $p=\frac{1}{2}$ \begin{align*} \sum_{d=k}^{n} \binom{d}{k} \left(\frac{1}{2}\right)^{d}\left(1-\frac{1}{2}\right)^{n-d}=\frac{1}{2^n}\binom{n+1}{k+1} \end{align*}

We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. So, we can write e.g. $$\binom{d}{k}=[z^k](1+z)^d$$

We start with general $p$, use the shorthand $q:=\frac{p}{1-p}$ and we observe: \begin{align*} \sum_{d=k}^{n}\binom{d}{k}p^d(1-p)^{n-d}&=(1-p)^n\sum_{d=k}^n\binom{d}{k}\left(\frac{p}{1-p}\right)^d\\ &=(1-p)^n\sum_{d=k}^{n}[z^k](1+z)^dq^d\\ &=(1-p)^n[z^k]\sum_{d=k}^{n}\left((1+z)q\right)^d\\ &=(1-p)^n[z^k]\left(\sum_{d=0}^{n}\left((1+z)q\right)^d-\sum_{d=0}^{k-1}\left((1+z)q\right)^d\right)\tag{1}\\ &=(1-p)^n[z^k]\left(\frac{1-\left((1+z)q\right)^{n+1}}{1-(1+z)q} -\frac{1-\left((1+z)q\right)^{k}}{1-(1+z)q}\right)\\ &=(1-p)^n[z^k]\frac{\left((1+z)q\right)^{k}-\left((1+z)q\right)^{n+1}}{1-(1+z)q}\tag{2} \end{align*}

Comment:

• In (1) we apply the formula for finite geometric series

• in (2) we could try to go on by expanding the denominator as series $$\frac{1}{1-(1+z)q}=\sum_{l\geq 0}\left((1+z)q\right)^l$$ and extracting via $[z^k]$ the coefficent of $z^k$. Regrettably, when doing so we will finally come back to the expression where we've started. But at least for the special case $p=\frac{1}{2}$ we can proceed.

We obtain continuing from (2) the

Special case: $p=\frac{1}{2},q=\frac{p}{1-p}=1$

\begin{align*} \frac{1}{2^n}\sum_{d=k}^n\binom{d}{k}&=\frac{1}{2^n}[z^k]\frac{(1+z)^{k}-(1+z)^{n+1}}{-z}\\ &=\frac{1}{2^n}[z^{k+1}]\left((1+z)^{n+1}-(1+z)^{k}\right)\\ &=\frac{1}{2^n}\left(\binom{n+1}{k+1}-\binom{k}{k+1}\right)\\ &=\frac{1}{2^n}\binom{n+1}{k+1} \end{align*}

and conclude

$$\sum_{d=k}^n\binom{d}{k}=\binom{n+1}{k+1}\qquad\qquad 0 \leq k \leq n$$