How to compute the following formula? The question is,
S wants to wants to broadcast messages to A and B until both A and B receive at least K messages. Everytime S broadcast a message, A has a probability of $p_1$ to receive it and B has a probability of $p_2$ to receive it. Links S-A and S-B are assumed to be independent. Then what is the expectation of number of broadcast S needs to make sure both A and B receive at least K messages.
In my derivation, it is computed as the following,
$
\begin{equation}
\begin{array}{lcl}
E & = & \sum_{x=K}^{\infty} x \cdot p(x)\\
\\
    & = & \sum_{x=K}^{\infty} x \cdot\{ {{x-1}\choose{K-1}} p_1^{K-1}(1-p_1)^{x-K}p_1{{x} \choose {K}} p_2^{K}\\
    \\
    &    & + {{x-1}\choose{K-1}} p_2^{K-1}(1-p_2)^{x-K}p_2{{x} \choose {K}} p_1^{K}\\
    \\
    &    & - {{x-1}\choose{K-1}} p_1^{K-1}(1-p_1)^{x-K}p_1 {{x-1}\choose{K-1}} p_2^{K-1}(1-p_2)^{x-K}p_2\}\\
    \\
    & = & \sum_{x=K}^{\infty} \{ {{x-1}\choose{K-1}} x  {{x} \choose {K}} p_1^K p_2^K (1-p_1)^{x-K}\\
    \\
    &    & + {{x-1}\choose{K-1}} x  {{x} \choose {K}} p_1^K p_2^K (1-p_2)^{x-K}\\
    \\
    &    & - {{x-1}\choose{K-1}} x  {{x-1} \choose {K-1}} p_1^K p_2^K (1-p_1)^{x-K} (1-p_2)^{x-K}\}\\
    \\
    & = & \sum_{x=K}^{\infty} \{ K{{x}\choose{K}} {{x} \choose {K}} p_1^K p_2^K (1-p_1)^{x-K}\\
    \\
    &    & + K{{x}\choose{K}} {{x} \choose {K}} p_1^K p_2^K (1-p_2)^{x-K}\\
    \\
    &    & - K{{x}\choose{K}} {{x-1} \choose {K-1}} p_1^K p_2^K (1-p_1)^{x-K} (1-p_2)^{K-k}\}\\
    \\
    & = & K p_1^K p_2^K \sum_{x=K}^{\infty} \{ {{x}\choose{K}} {{x} \choose {K}}  (1-p_1)^{x-K}
    + {{x}\choose{K}} {{x} \choose {K}} (1-p_2)^{x-K}\\
    \\
    &    & - {{x}\choose{K}} {{x-1} \choose {K-1}}  (1-p_1)^{x-K} (1-p_2)^{x-K}\}\\
\end{array}
\end{equation}
$
How can I make this formula simpler? Thanks a lot.
Sorry about the misunderstanding. My goal is to compute the expectation of minimal number of broadcasts such that both A and B receive at least K messages.
 A: Let $X_n$ denote the state of the system after $n$ broadcasts encoding $(k_A,k_B)$, where $k_A$ and $k_B$ are number messages received by $A$ and $B$ respectively.
The termination condition is 
$$
    \Omega = \{ (k_A,k_B) \colon \min(k_A, k_B) = k \}
$$
Let $T = \inf\{n \colon X_n \in \Omega, X_{n-1} \not\in \Omega \}$. We are interested in determining $\mathbb{E}(T)$.
The probability that $T$ equals $n$ is computed conditioning on $\Omega \not\ni X_{n-1} \to X_n \in \Omega$:
$$ \begin{eqnarray}
   \mathbb{P}\left(T=n\right) &=& \mathbb{P}(K_A=k-1)\mathbb{P}(K_B \geqslant k) p_1 \\ &\phantom{=}& + \mathbb{P}(K_B=k-1)\mathbb{P}(K_A \geqslant k) p_2\\ &\phantom{=}& + 
   \mathbb{P}(K_A=k-1) \mathbb{P}(K_B=k-1) p_1 p_2 \\ 
   &=& \binom{n-1}{k-1} p_1^{k} q_1^{n-k} \sum_{m=k}^{n-1} \binom{n-1}{m} p_2^m q_2^{n-1-m} \\ &+&  \binom{n-1}{k-1} p_2^{k} q_2^{n-k} \sum_{m=k}^{n-1} \binom{n-1}{m} p_1^m q_1^{n-1-m} \\
   &+& \binom{n-1}{k-1} p_2^{k} q_2^{n-k} \binom{n-1}{k-1} p_1^{k} q_1^{n-k}
\end{eqnarray}
$$
I am not able to compute the expected value in the closed form, however, I am able to use Mathematica to find means for specific values of $k$:
pdf[{k_Integer, p1_, p2_}, n_] := 
 FullSimplify@
  FunctionExpand@
   PiecewiseExpand[
    Refine[p1 PDF[BinomialDistribution[n - 1, p1], 
        k - 1] SurvivalFunction[BinomialDistribution[n - 1, p2], 
        k - 1], k \[Element] Integers] + 
     Refine[p2 PDF[BinomialDistribution[n - 1, p2], 
        k - 1] SurvivalFunction[BinomialDistribution[n - 1, p1], 
        k - 1], k \[Element] Integers] + 
     Refine[p1 p2 PDF[BinomialDistribution[n - 1, p2], k - 1] PDF[
        BinomialDistribution[n - 1, p1], k - 1], 
      k \[Element] Integers]]

The density can be computed in closed form for explicit values of $k$, and Sum can be used to find the mean:
In[61]:= With[{k = 1},
 Sum[n Assuming[n >= k, 
    FullSimplify@Refine@FunctionExpand[pdf[{k, p1, p2}, n]]], {n, 
   k, \[Infinity]}, Assumptions -> 0 < p1 < 1 && 0 < p2 < 1]
 ]

Out[61]= (-p1^2 - p1 p2 + p1^2 p2 - p2^2 + 
 p1 p2^2)/(p1 p2 (-p1 - p2 + p1 p2))

