Let $F\left( n \right) = \sum\limits_{k = 0}^n {{{\left( {C_n^k{p^k}{{\left( {1 - p} \right)}^{n - k}}} \right)}^2}} $, prove $F\left( n \right) \ge F\left( {n + 1} \right)$.
UPDATE: More general, denote $F\left( n \right) = \sum\limits_{k = 0}^n {C_n^kp_1^kq_1^{n - k}C_n^kp_2^kq_2^{n - k}}$, where ${q_1} = 1 - {p_1}$ and ${q_2} = 1 - {p_2}$, prove $F\left( n \right) \ge F\left( {n + 1} \right)$.
$\color{red}{^\bf{{\rm{New}}}}$ UPDATE 2: More general, denote $F\left( n \right) = \sum\limits_{k = 0}^n {C_n^kp_1^kq_1^{n - k}\sum\limits_{i = 0}^k {C_k^ip_2^iq_2^{k - i}C_{n - k}^{k - i}p_3^{k - i}q_3^{\left( {n - k} \right) - \left( {k - i} \right)}} }$, where $q_1=1-p_1$, $q_2=1-p_2$, and $q_3=1-p_3$, is it true that $F\left( n \right) \ge F\left( {n + 1} \right)$?
As we can see, when $p_2=p_3$, according to Vandermonde's identity, there is $\sum\limits_{i = 0}^k {C_k^iC_{n - k}^{k - i}} = C_n^k$, $\sum\limits_{i = 0}^k {C_k^ip_2^iq_2^{k - i}C_{n - k}^{k - i}p_3^{k - i}q_3^{\left( {n - k} \right) - \left( {k - i} \right)}} = C_n^kp_2^kq_2^{n - k}$, and thus the problem becomes the last updated one. Mr. Petrov has provided an elegant proof and he also comes up with an brilliant idea to compute $F\left(n\right)$ to solve the form when $p_2=p_3$. It would be great that you can prove this form ($p_2≠p_3$) and at the same time kindly provide an efficient method to compute $F\left(n\right)$.
