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.
 A: How to find the numerically greatest term (NGT) in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$?
$$(3x+5y)^{12}=(3x)^{12}\left(1+\frac{5y}{3x}\right)^{12}$$
When compared $\left(1+\frac{5y}{3x}\right)^{12}$ with $(1+x)^n$, we got, $n=12$, a positive integer, and $x=\left(\frac53\right)\left(\frac yx\right)=\left(\frac53\right)\left(\frac{4/3}{1/2}\right) =\frac{40}{9}$. Thus, if rth-term $T_r$ is the numerically greatest term, $\left(T_{r+1}/T_r\right) \lt 1$.
$$T_{r+1}=~^nC_r(|x|)^r=\frac{n!}{(n-r)!r!}(|x|)^r ~\text{and} ~T_{r}=~^nC_{r-1}(|x|)^{r-1}=\frac{n!}{(n-r+1)!r-1!}(|x|)^{r-1}$$
$$\text{Thus,}~\left(\frac{T_{r+1}}{T_r}\right)=\left(\frac{n-r+1}{r}\right)|x| \lt 1 ~\text{only when,} $$
$$\frac{(n+1)|x|}{1+|x|}\lt r$$
$$\frac{(n+1)|x|}{1+|x|}=\frac{(12+1)\left(\frac{40}{9}\right)}{1+\frac {40}{9}}=\frac {13\cdot 40}{49}=10\frac {30}{49} \lt r$$
Note that if $\frac{(n+1)|x|}{1+|x|}$ is an integer, then both $T_r$ and $T_{r+1}$ are numerically greatest terms (that's the reason I derive commonly used unequality to show you).  
Thus, considering $(1+x)^n$ expression,
 $$T_r=T_{11}=~^{12}C_{10}\left(\frac{40}{9}\right)^{10}=\frac{12!}{(13-11)!10!}\left(\frac{40}{9}\right)^{10}=66\cdot \left(\frac{40}{9}\right)^{10}$$
Therefore the complete NGT is:
$$\text{NGT}=(3x)^{12}\cdot 66\cdot \left(\frac{40}{9}\right)^{10}= \left(\frac{3}{2}\right)^{12}\cdot 66\cdot \left(\frac{40}{9}\right)^{10}$$.
A: 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}
