Question

Given that the $4th$ term in the expansion of ${(2+\frac{3x}{8})}^{10}$ has the maximum numerical value, find the range of values of $x$ for which this will be true.

Answer 1

Let $T_4$ be numerically the greatest term in the expansion of $2^{10}{(1+\frac{3x}{16})}^{10}$
$\therefore \left|\frac{T_4}{T_3}\right|\ge 1$ and $\left|\frac{T_4}{T_5}\right|\ge 1$ $\Rightarrow \left|\frac{T_4}{T_3}\right|\ge 1$ and $\left|\frac{T_5}{T_4}\right|\le 1$
Now, $\frac{Tr+1}{Tr}=\frac{n-r+1}{r}x.$
Taking $r=3$ and $r=4,$ replacing $x$ by $\frac{3x}{16}$ and $n=10,$ we get
$|\frac{11-3}{3}-\frac{3x}{16}|\ge 1$ and $|\frac{11-4}{4}-\frac{3x}{16}|\le 1$
$\Rightarrow \left|x\right|\ge 2$ and $\left|x\right|\le \frac{64}{21}$ ...(1)
$\therefore x^2\ge 4$ and $x^2\le {\left(\frac{64}{21}\right)}^2$
$\Rightarrow 4\le x^2\le {\left(\frac{64}{21}\right)}^2$
$\Rightarrow 2\le x\le \frac{64}{21}$ and $\frac{-64}{21}\le x\le -2$