Question

The ratio of the $5^{th}$ term from the beginning to the $5^{th}$ term from the end in the binomial expansion of ${(2^{\frac{1}{3}}+\frac{1}{2(2{)}^{\frac{1}{3}}})}^{10}$ is:

• A. $1:2(6{)}^{\frac{1}{3}}$
• B. $2(36{)}^{\frac{1}{3}}:1$
• C. $4(36{)}^{\frac{1}{3}}:1$
• D. $1:4(16{)}^{\frac{1}{3}}$

Answer 1

The correct option is C $4(36{)}^{\frac{1}{3}}:1$

${(2^{\frac{1}{3}}+\frac{1}{2(3{)}^{\frac{1}{3}}})}^{10}$
We know that $5^{th}$ term from end will be $7^{th}$ term from begining,
$\therefore \frac{T_5}{T_7}=\frac{{}^{10}{C}_4\cdot {\left(2^{\frac{1}{3}}\right)}^6\cdot {\left(\frac{1}{2.(3{)}^{\frac{1}{3}}}\right)}^4}{{}^{10}{C}_6\cdot {\left(2^{\frac{1}{3}}\right)}^4.{\left(\frac{1}{2\cdot (3{)}^{\frac{1}{3}}}\right)}^6}$
$=\frac{{\left(2^{\frac{1}{3}}\right)}^2}{{\left(\frac{1}{2\cdot (3{)}^{\frac{1}{3}}}\right)}^2}$
$=\frac{2^{\frac{2}{3}}\cdot 2^2\cdot 3^{\frac{2}{3}}}{1}$
$\frac{T_5}{T_7}=\frac{4\cdot (36{)}^{\frac{1}{3}}}{1}$
$T_5:T_7=4\cdot (36{)}^{\frac{1}{3}}:1$