Question

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

Answer 1

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

7th term from beginning is ${}^n{C}_6(\sqrt[3]{32}{)}^{n-6}{\left(\frac{1}{\sqrt[3]{33}}\right)}^{6}7th$ term from end is

${}^n{C}_{n-6}(\sqrt[3]{32}{)}^6{\left(\frac{1}{3\sqrt{3}}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{33}}\right)}^6$

Given $\frac{\text{7th term from beginning}}{\text{7th term from end}}$

$=\frac{{(}^n{C}_6\left)\right(\sqrt[3]{32}{)}^{n-12}(\frac{1}{\sqrt[3]{33}}{)}^{12-n}}{{(}^n{C}_{n-6})}$

$=\frac{{(}^n{C}_6\left)\right(\sqrt[3]{32}{)}^{n-12}(\sqrt[3]{33}{)}^{n-12}}{{(}^n{C}_{n-6})}$

$=\frac{{(}^n{C}_6\left)\right(6{)}^{\frac{n-12}{3}}}{{(}^n{C}_{n-6})}=\frac{1}{6}$

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

n=12 -3=9