Question

If the ratio of the seventh term from the beginning of the binomial expansion of ${(2^{1/3}+\frac{1}{3^{1/3}})}^x$ to the seventh term from its end is $\frac{1}{6}$, then the value $x$ is

• A. $5$
• B. $11$
• C. $9$
• D. $7$

Answer 1

The correct option is C $9$

We have
${(2^{1/3}+\frac{1}{3^{1/3}})}^x$
$\because$ 7th term of expansion from beginning
$={{}^{x}C}_6{\left(2^{1/3}\right)}^{x-6}{\left(\frac{1}{3^{1/3}}\right)}^6$
And 7th term of expansion from end
$={{}^{x}C}_{x-6}{\left(2^{1/3}\right)}^6{\left(\frac{1}{3^{1/3}}\right)}^{x-6}$
Given :
$\frac{7thtermexpansionfrombeginning}{7thtermexpansionfromend}=\frac{1}{6}$
$\Rightarrow \frac{{{}^{x}C}_6{\left(2^{1/3}\right)}^{x-6}{\left(\frac{1}{3^{1/3}}\right)}^6}{{{}^{x}C}_{x-6}{\left(2^{1/3}\right)}^6{\left(\frac{1}{3^{1/3}}\right)}^{x-6}}=\frac{1}{6}$
$\Rightarrow \frac{{{}^{x}C}_6{\left(2^{1/3}\right)}^{x-12}}{{{}^{x}C}_6{\left(\frac{1}{3^{1/3}}\right)}^{x-12}}=\frac{1}{6}$
${\left(6^{1/3}\right)}^{x-12}=\frac{1}{6}$
$\Rightarrow {\left(\frac{1}{6}\right)}^{-1/3(x-12)}=\frac{1}{6}$
$\therefore -\frac{1}{3}(x-12)=1$
$12-x=3$
$\Rightarrow$ $x=9$