Find the value of r in the largest term in $(5+2x)^{14}$ where $T(r)$ is the $r$-th term and $x=4/3$

Question

So I've got the answer as $r=5$. Here's how I got it.

$$T(r) = \binom{14}{r}\cdot(5)^{14-r}\cdot(2x)^r$$

$$T(r+1) = \binom{14}{r+1}\cdot(5)^{13-r}\cdot(2x)^{r+1}$$

$$T(r+1)/T(r) > 1$$

$$r > 4.21$$

So that $$r=5$$

My question is why can't we take $r=6$ or $r=7$?

Answer 1

I prefer to go the other way.

$$T_r = \binom{14}{r}5^r \left[ ~\frac{8}{3} ~\right]^{14-r}.$$

As $~r~$ goes from $~0~$ to $~7,~$ you have that $~\displaystyle \binom{14}{r}~$ is strictly increasing. Then, since $~5 > \dfrac{8}{3}~$ you must have that $~T_r~$ is strictly increasing.

So, the optimal value of $~r~$ must be $~\geq 7.~$

For $~r \in \{7,8,\cdots,13\}~$ you have that

$$\frac{\binom{14}{r+1}}{\binom{14}{r}} = \frac{14-r}{r+1}.$$

Therefore, for $~r \in \{7,8,\cdots,13\},~$ you have that

$$\frac{T_{r+1}}{T_r} = \frac{14-r}{r+1} \times \frac{5}{(8/3)} = \frac{14-r}{r+1} \times \frac{15}{8}.$$

For $~r \in \{7,\cdots,13\},~$

• Let $~f(r)~$ denote $~\dfrac{T_{r+1}}{T_r}.$

• Let $~g(r)~$ denote $~\dfrac{T_r}{T_7}.$
  This implies that for $~r \geq 8,~$ that
  $g(r) = f(7) \times f(8) \times \cdots \times f(r-1).$

The goal is to maximize $~g(r).$

Then:

• $\displaystyle f(7) = \frac{7}{8} \times \frac{15}{8} > 1.$

• $\displaystyle f(8) = \frac{6}{9} \times \frac{15}{8} = \frac{90}{72} > 1.$

• $\displaystyle f(9) = \frac{5}{10} \times \frac{15}{8} = \frac{75}{80} < 1.$

At this point, the analysis can stop, because for $~r \in \{9,10,11,12,13\},~$ the fraction $~\dfrac{14-r}{r+1}~$ is strictly decreasing.

Therefore, you have that $~f(7) > 1, ~f(8) > 1,~$ and $f(r) < 1 ~: ~r \in \{9,10,11,12,13\}.$

Therefore, $~g(r)~$ is maximized at $~g(9) = f(7) \times f(8).$

Therefore, $~T_9~$ is the maximum coefficient.

Noting that my definition of $~T_9~$ is equivalent to the original poster's definition of $~T(5),~$ my answer agrees with the original poster's answer.

Answer 2

$T(r+1)/T(r) > 1$ is not true for all $r$.
However, your idea is correct, we'll find $T(r+1)/T(r)$: $$\frac{T(r+1)}{T(r)} = \frac 8{15} \cdot \frac{14-r}{r+1}$$ Set $\color{green}{T(r+1)/T(r)>1}$: $$\frac 8{15} \cdot \frac{14-r}{r+1}>1 \iff 112 - 8r>15r+15$$ which gives $$\color{green}{r<97/23 \approx 4.217}$$ Now, set $\color{blue}{T(r+1)/T(r)<1}$: $$\frac 8{15} \cdot \frac{14-r}{r+1}<1 \iff 112 - 8r<15r+15$$ which gives $$\color{blue}{r>97/23 \approx 4.217}$$ Since $r$ is an integer, we have that, when $r = 4$, $T(r+1) = T(5)$ is greater than $T(4)$ using the green inequality. And using the blue one, when $r = 5$, $T(r+1) = T(5)$ is greater than $T(6)$.
So, $T(r)$ increases from $r = 0$ till $5$ and decreases from $r = 5$ till $14$. Hence, the maximum term is $T(5)$.