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# A ratio of the 5th term from the beginning

Question:

A ratio of the $5^{\text {th }}$ term from the beginning to the $5^{\text {th }}$ term from the end in the binomial

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

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

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

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

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

Correct Option: , 4

Solution:

$\frac{\mathrm{T}_{5}}{\mathrm{~T}_{5}^{1}}=\frac{{ }^{10} \mathrm{C}_{4}\left(2^{1 / 3}\right)^{10-4}\left(\frac{1}{2(3)^{1 / 3}}\right)^{4}}{{ }^{10} \mathrm{C}_{4}\left(\frac{1}{2\left(3^{1 / 3}\right)}\right)^{10-4}\left(2^{1 / 3}\right)^{4}}=4 \cdot(36)^{1 / 3}$