# A ratio of the 5th term from the begining to the 5 th term from

Question:

A ratio of the $5^{\text {th }}$ term from the begining to the 5 th term from

the end in the binomial expansion of $\left(2^{1 / 3}+\frac{1}{2(3)^{1 / 3}}\right)^{10}$ is:

1. (1) $1: 2(6)^{\frac{1}{3}}$

2. (2) $1: 4(16)^{\frac{1}{3}}$

3. (3) $4(36)^{\frac{1}{3}}: 1$

4. (4) $2(36)^{\frac{1}{3}}: 1$

Correct Option: , 2, 3

Solution:

$\left(2^{\frac{1}{3}}+\frac{1}{2(3)^{\frac{1}{3}}}\right)^{10}={ }^{10} C_{0}\left(2^{\frac{1}{3}}\right)^{0}\left(\frac{1}{2(3)^{1 / 3}}\right)^{10}+$

$\cdots+{ }^{10} C_{10}\left(2^{\frac{1}{3}}\right)^{10}\left(\frac{1}{2(3)^{1 / 3}}\right)^{0}$

$5^{\text {th }}$ term from beginning $T_{5}={ }^{10} C_{4}\left(2^{\frac{1}{3}}\right)^{6} \frac{1}{\left(2.3^{\frac{1}{3}}\right)^{4}}$

and $5^{\text {th }}$ term from end $T_{11-5+1}={ }^{10} C_{6}\left(2^{\frac{1}{3}}\right)^{4}\left(\frac{1}{2.3^{\frac{1}{3}}}\right)^{6}$

$\therefore \quad T_{5} \cdot T_{7}=$

${ }^{10} C_{4}\left(2^{\frac{1}{3}}\right)^{6}\left(\frac{1}{2.3^{\frac{1}{3}}}\right)^{4}:{ }^{10} C_{6}\left(2^{\frac{1}{3}}\right)^{4}\left(\frac{1}{2.3^{\frac{1}{3}}}\right)^{6}$

$=\left(2^{\frac{1}{3}}\right)^{2}:\left(\frac{1}{2.3^{\frac{1}{3}}}\right)^{2}$

$=\frac{2^{\frac{2}{3}} \cdot 2^{2} \cdot 3^{\frac{2}{3}}}{1}=4(6)^{\frac{2}{3}}: 1=4 \cdot(36)^{\frac{1}{3}}: 1$