Find $R$ such that $\sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(R)}^n\cdot{(1-R)}^{3k-n}$ is constant for all $k\in\mathbb{N}$ Given $A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(R)}^n\cdot{(1-R)}^{3k-n}$ with $0<R<1$.
The sequence $A_k$ seems to be decreasing for $R\leq0.6$ and increasing for $R\geq0.8$.
How can we analytically find the value of $R$ for which the sequence $A_k$ is constant for all $k\in\mathbb{N}$?
 A: For any value of $k\in\mathbb{N}$ we have that $A_k(R)$ is a polynomial with degree $3k$ that takes positive values over $(0,1)$ and has a root of multiplicity $2k$ in $0$. Despite the fact that the polynomials $A_1,A_2,A_3$ assume very close values over $[0.7,0.8]$ (as depicted below), there is no $R$ such that $A_1(R)=A_2(R)=A_3(R)$, so the sequence $\{A_k\}_{k\in\mathbb{N}}$ cannot be constant.


However, by the law of large numbers,
$$\lim_{k\to +\infty}A_k(2/3) = 1/2,$$
hence $\frac{2}{3}$ is a kind of critical value. It happens that $A_k(2/3)$ decreases towards $\frac{1}{2}$ as $k$ increases, and:
$$ A_k(2/3)=\frac{1}{2}+O\left(\frac{1}{\sqrt{k}}\right).$$
For values of $R$ bigger than $2/3$, $A_k(R)$ regarded as a function of $k$ first decreases, then increases.
A: First you should proof, if there is a constant value R, which fullfill the condition.
Just solve the equation $A_2=A_3$ and calculate $R^*$. Then calculate $A_k(R^*)$ for some $k \in  \mathbb N \backslash \{ 2,3\}$ and compare the values.
