Question

In the binomial expansion of $(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}}{)}^n$, the ratio of the $7th$ term from the beginning to the $7th$ term from the end is $1:6$. Find $n$.

• A. $9$
• B. $7$
• C. $5$
• D. $3$

Answer 1

Answer: (A)

$9$

$(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}}{)}^n$

Here $a=\sqrt[3]{32}$ $b=\frac{1}{\sqrt[3]{33}}$

We have the general form for term in binomial expression,

$T_{r+1}={}_r^n\text{C}{a}^{n-r}{b}^r$

Given, $\frac{7thtermfrombeginning}{7thtermfromtheend}=\frac{1}{6}$

$\frac{{}_6^n\text{C}(2^{\frac{1}{3}}{)}^{n-6}.{\left(\frac{1}{3^{\frac{1}{3}}}\right)}^6}{{}_{n-6}^n\text{C}(2^{\frac{1}{3}}{)}^6.{\left(\frac{1}{3^{\frac{1}{3}}}\right)}^{n-6}}=\frac{1}{6}$

$⟹\frac{(2^{\frac{1}{3}}{)}^{n-12}}{{\left(\frac{1}{3^{\frac{1}{3}}}\right)}^{n-12}}=\frac{1}{6}$

$2^{\frac{n-12}{3}}\times 3^{\frac{n-12}{3}}=\frac{1}{6}$

$6^{\frac{n-12}{3}}=6^{-1}$

$n-12=-3$

$n=9$