Prove by induction that $A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$ is decreasing I want to prove that the following sequence is monotonously decreasing:
$A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$
I think it should be pretty easy to do so by induction:


*

*$A_1 = \binom{3}{2}\cdot{(\frac{60}{100})}^2\cdot{(\frac{40}{100})}^1+\binom{3}{3}\cdot{(\frac{60}{100})}^3\cdot{(\frac{40}{100})}^0 = 0.648$

*$A_2 = \binom{6}{4}\cdot{(\frac{60}{100})}^4\cdot{(\frac{40}{100})}^2+\binom{6}{5}\cdot{(\frac{60}{100})}^5\cdot{(\frac{40}{100})}^1+\binom{6}{6}\cdot{(\frac{60}{100})}^6\cdot{(\frac{40}{100})}^0 = 0.54432$

*Let's assume that $A_{k} > A_{k-1}$

*Let's prove that $A_{k+1} > A_{k}$
So how do I go about with that? Or is there another way?

It happens that
$$ A_k = \left(\frac{18}{125}\right)^k\binom{3k}{2k}\phantom{}_2 F_1\left(1,-k;1+2k;-\frac{3}{2}\right),$$
and a clever idea is to use the Stirling approximation together with the Gauss continued fraction for the hypergeometric function in order to give tight bounds for $A_k$, then prove $A_{k+1}<A_{k}$ for any $k$ big enough.
$$\phantom{}_2 F_1\left(1,-k;1+2k;-\frac{3}{2}\right)=1+\frac{\frac{3k}{2+4k}}{1-\frac{\frac{3k-3}{4+4k}}{1+\frac{3k-3}{4+4k}+\frac{\frac{3(2-k)}{2(3+2k)}}{1+\ldots}}}\in\left[4-\frac{60}{k+20},4-\frac{60}{k+53}\right],$$
so the asymptotic behaviour of $A_k$ is:
$$A_k=\left(\frac{243}{250}\right)^k\sqrt{\frac{12}{\pi k}}\left(1+\Theta\left(\frac{1}{k}\right)\right)$$
and $\{A_k\}_{k\in\mathbb{N}}$ decreases for sure from a certain point on. If we compute the constants hidden in the $\Theta$-notation, we can also notice that 
$$\left(\frac{250}{243}\right)^k A_k$$
is a decreasing function, hence $A_k$ is a decreasing function, too.
