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**: 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 $F\left( n \right) \ge F\left( {n + 1} \right)$?



How to prove the sum of n squared binomial probabilities does not increase as n increases
Denote $q=1-p$, write $[x^a]f(x)$ for a coefficient of $x^a$ in the polynomial (or Laurent polynomial) $f(x)$. We have $$F(n)=[1](px+q)^n(px^{-1}+q)^n=[1]((p^2+q^2)+qp(x+x^{-1}))^n=\\=(2\pi)^{-1}\int\_{0}^{2\pi}((p^2+q^2)+qp(e^{it}+e^{-it}))^ndt.$$
The function $(p^2+q^2)+qp(e^{it}+e^{-it})=p^2+q^2+2pq\cos t$ is non-negative and its values do not exceed 1, thus the result.


I use the useful relation $[1]\varphi(x)=(2\pi)^{-1}\int\_0^{2\pi} \varphi(e^{it})dt$ for any Laurent polynomial $\varphi$. 


UPD. For the updated problem, use the same integral representation for $p=\sqrt{p\_1p\_2},q=\sqrt{q\_1q\_2} $. We need to check that $p+q\leqslant 1$, this follows from $p\leqslant (p\_1+p\_2)/2, $  

$q\leqslant (q\_1+q\_2)/2. $


$$
\sum\limits\_{i = 0}^k {C\_k^ip\_2^iq\_2^{k - i}C\_{n - k}^{k - i}p\_3^{k - i}q\_3^{\left( {n - k} \right) - \left( {k - i} \right)}}=[t^k](tp\_2+q\_2)^k(tp\_3+q\_3)^{n-k}
$$


About updated function $F(n)$. Denoting $h(z)=z^{-1}p\_1(zp\_2+q\_2)+q\_1(zp\_3+q\_3)$ we have $F(n)=[1](h(z))^n$, thus
$$
F(n)=\frac1{2\pi i}\int h(z)^ndz/z,$$
where integral is taken by any circuit around 0. It is natural to take the circuit with real value of $h(z)=p\_1p\_2+q\_1q\_3+q\_1p\_3 z+p\_1q\_2 z^{-1}$. The natural choice is the contour $z=\sqrt{\frac{p\_1q\_2}{q\_1p\_3}}e^{i\theta}$, we get $h(z)=p\_1p\_2+q\_1q\_3+2\sqrt{p\_1q\_1q\_2p\_3}\cos \theta$. We have $|h(z)|\leqslant p\_1p\_2+q\_1q\_3+p\_1q\_2+q\_1p\_3=p\_1+q\_1=1$, thus everything works nicely if we have $h(z)\geqslant 0$. This is not always the case. For example, if $p\_2=q\_3=0$ we get $h(z)=2\sqrt{p\_1q\_1}\cos \theta$ and the integral equals 0 for odd values of $n$ and is positive for even values of $n$. Needless to say, this could be seen from the very definition of $F(n)$, but for what it worth I leave here the general explanation.


