# 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 $k\in\mathbb{N}$?

• I guess it's a trivial suggestion, but did you look at $$1=(R+(1-R))^{3k}=\sum_{n=0}^{3k}\binom{3k}{n}R^n(1-R)^{3k-n}$$ Then you can denote $$B_k=\sum_{n=k}^{2k}\binom{3k}{n}R^n(1-R)^{3k-n}, \hspace{5pt}C_k=\sum_{n=0}^{k}\binom{3k}{n}R^n(1-R)^{3k-n}$$ And then $A_k+B_k+C_k=1$. I'm not sure it leads anywhere, but seems like a good place to start from. $A_k$ and $C_k$ look symmetric in a way. – Dennis Gulko Jul 17 '14 at 18:16
• For a given value of $R$, how can the derivative of your expression with respect to $k$ vanish? – Jack D'Aurizio Jul 17 '14 at 19:10

## 2 Answers

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.

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.