Question

In the expansion of ${(\frac{1}{2}+\frac{2x}{3})}^n$ when $x=-\frac{1}{2}$, it is known that 3rd term is the greatest term. Find possible integral values of $n$.

Answer 1

$x=-\frac{1}{2}$

So third term is $(\frac{n}{2})$ ${\left(\frac{1}{2}\right)}^2{\left(\frac{-1}{3}\right)}^{n-2}$

Now for it to be greatest it must be greater than the second term and the fourth term.

Second term is $(\frac{n}{1})$ $\left(\frac{1}{2}\right){\left(\frac{-1}{3}\right)}^{n-1}$

$\frac{T_3}{T_2}>1$

$\Rightarrow \frac{n-1}{2}\times \frac{\frac{1}{2}}{-\frac{1}{3}}>1$

$\Rightarrow n>\frac{-1}{3}$

$T_4=$$(\frac{n}{3})$ ${\left(\frac{1}{2}\right)}^3{\left(\frac{-1}{3}\right)}^{n-3}$

$\frac{T_4}{T_3}<1$

$\Rightarrow \frac{2(n-2)}{3}\frac{\frac{1}{2}}{-\frac{1}{3}}<1$

$\Rightarrow n<1$

Hence, no values of $n$ for which third term is greatest.