<s>[INST] How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$ How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$?
My attempt
$$(3x+5y)^{12}=\left(3x\left(1+\frac{5y}{3x}\right)\right)^{12}$$
$$=3^{12}x^{12}\left(1+\frac{5y}{3x}\right)^{12}$$
I then compared $\left(1+\frac{5y}{3x}\right)^{12}$ with $(1+x)^n$ and got $n=17,\space x=(\frac53)(\frac yx)=(\frac53)\frac {\frac43}{\frac12} =\frac{40}{9}$.
Then I used $$\frac{(n+1)|x|}{1+|x|}=\frac{(12+1)\left(\frac{40}{9}\right)}{1+\frac {40}{9}}$$
After solving, I got
$$=10\frac{30}{49}$$ which is not an integer.
I am stuck here, can anyone explain how to solve this problem.
 [/INST] The $n^{th}$ term of $\left(\frac 32 + \frac{20}{3} \right)^{12}$ is 
$t_n =\binom{12}{n}\left(\frac 32\right)^{12-n}\left(\frac{20}{3}\right)^n$ for $n = 0,1,\dots,12$. The ratio of consecutive terms is 
$\rho_n = \dfrac{t_{n+1}}{t_n} = \dfrac{40}{9}\left(\dfrac{12-n}{n+1}\right)$.
These ratios are easy to compute
\begin{array}{|c|cccccccccccc|}
\hline
\text{n} & 0 & 1 & 2  & 3 
         & 4 & 5 & 6  & 7
         & 8 & 9 & 10 & 11\\
\hline
       r_n & \frac{160}{3}  & \frac{220}{9} & \frac{400}{27} & 10 
           & \frac{64}{9}   & \frac{140}{27} & \frac{80}{21}  & \frac{25}{9}
           & \frac{160}{81} & \frac{4}{3}   & \frac{80}{99}  & \frac{10}{27} \\
\hline
\end{array}
The last occurence of $r_n>1$ happens at $n=9$. So 
$t_{10} =\binom{12}{10}\left(\frac 32\right)^2\left(\frac{20}{3}\right)^{10}$ will be the largest term.
As @BarryCipra has pointed out. It is not necessary to compute the above table. We can solve the following inequality.
\begin{align}
   \dfrac{t_{n+1}}{t_n} &\ge 1 \\
   \dfrac{40}{9}\left(\dfrac{12-n}{n+1}\right) &\ge 1 \\
   480-40n &\ge 9n + 9 \\
   49n &\le 472 \\
   n &\le \dfrac{472}{49} \\
   n &\le 9 \\
\end{align}
</s>