How to find the numerically greatest term in the expansion of $(3x+5y)^{12}$ when $x=\frac12,y=\frac43$

Question

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.

Answer 1

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}

Answer 2

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: $$\text{NGT}=(3x)^{12}\cdot 66\cdot \left(\frac{40}{9}\right)^{10}= \left(\frac{3}{2}\right)^{12}\cdot 66\cdot \left(\frac{40}{9}\right)^{10}$$.