MHB Optimizing Binomial Coefficients for Maximum Value

[anemone]

From the binomial theorem, we have

$\displaystyle \begin{align*}\left(1+\dfrac{1}{5}\right)^{1000}&={1000 \choose 0}\left(\dfrac{1}{5}\right)^{0}+{1000 \choose 1}\left(\dfrac{1}{5}\right)^{1}+{1000 \choose 2}\left(\dfrac{1}{5}\right)^{2}+\cdots+{1000 \choose 1000}\left(\dfrac{1}{5}\right)^{1000}\\&=A_0+A_1+A_2+\cdots+A_{1000} \end{align*}$

where $\displaystyle A_k={1000 \choose k}\left(\dfrac{1}{5}\right)^{k}$.

For which $k$ is $A_k$ the largest?

 

[M R]

From the binomial theorem, we have

$\displaystyle \begin{align*}\left(1+\dfrac{1}{5}\right)^{1000}&={1000 \choose 0}\left(\dfrac{1}{5}\right)^{0}+{1000 \choose 1}\left(\dfrac{1}{5}\right)^{1}+{1000 \choose 2}\left(\dfrac{1}{5}\right)^{2}+\cdots+{1000 \choose 1000}\left(\dfrac{1}{5}\right)^{1000}\\&=A_0+A_1+A_2+\cdots+A_{1000} \end{align*}$

where $\displaystyle A_k={1000 \choose k}\left(\dfrac{1}{5}\right)^{k}$.

For which $k$ is $A_k$ the largest?

I looked for k such that $\frac{1000!}{k!(1000-k)!5^k}>\frac{1000!}{(k+1)!(999-k)!5^{k+1}}$.

This simplifies to $\frac{1}{1000-k}>\frac{1}{5k+5}$.

Leading to k=166 if I didn't make any dopey errors.

 

[anemone]

Thanks M R for your participation and your correct solution!(Yes)

A solution proposed by other:

Spoiler

Note that $\displaystyle A_k={1000 \choose k}\left(\dfrac{1}{5}\right)^k=\dfrac{1000!}{5^k(k!)(1000-k)!}$, so to maximize $A_K$ means we must minimize its denominator, and if we let it as $P_k=5^k(k!)(1000-k)!$, for all $k$, we have $P_{k+1}=5^{k+1}((k+1)!)(1000-(k+1))!=5^{k+1}((k+1)!)(999-k)!=P_k\left(\dfrac{5(k+1)}{1000-k}\right)$.

For small values of $k$, $\dfrac{5(k+1)}{1000-k}<1$, so we must have $P-0>P_1>P_2>\cdots$. The minimum value will thus come at the smallest value of $k$ for which $\dfrac{5(k+1)}{1000-k}>1$, so we must have $5(k+1)>1000-k\,\,\rightarrow k=166$