Question

If the sixth term from the beginning is equal to the sixth term from the end in the expansion of ${(2^{1/5}+3^{-1/5})}^n$, then the sum of the rational terms in the expansion is

• A. $\frac{1349}{9}$
• B. $\frac{1549}{9}$
• C. $\frac{1759}{9}$
• D. $\frac{1119}{9}$

Answer 1

The correct option is B $\frac{1549}{9}$

In the expansion of ${(2^{1/5}+3^{-1/5})}^n$,
$T_{r+1}={}^n{C}_r{\left(2^{1/5}\right)}^{n-r}{\left(3^{-1/5}\right)}^r$

Given that $T_6=T_{n-4}$
$\Rightarrow {}^n{C}_5\left(\frac{2^{n/5}}{6}\right)={}^n{C}_{n-5}\left(\frac{6}{3^{n/5}}\right)$
$\Rightarrow 6^{n/5}=36$
$\Rightarrow n=10$

$\therefore T_{r+1}={}^{10}{C}_r\cdot {\left(2\right)}^{\frac{10-r}{5}}\cdot {\left(3\right)}^{\frac{-r}{5}}$
The term will be rational only when $r=0,5,10$
$\therefore$ Sum of rational terms $=T_1+T_6+T_{11}$
$={}^{10}{C}_0\cdot 2^2+{}^{10}{C}_5\cdot \frac{2}{3}+{}^{10}{C}_{10}\cdot \frac{1}{9}$
$=4+168+\frac{1}{9}$
$=\frac{1549}{9}$