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:

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Binomial Theorem

A ratio of the

5^{th}

term from the beginning to the

5^{th}

term from the end in the binomial expansion of

(2^{1/3} + \frac{1}{2 ( 3 ) ^{1/3}})^{10}

is:

Answer: (C)

In the expansion of

(2^{1/3} + \frac{1}{2 \cdot 3 ^{1/3}})^{10}

there are

11

terms, so the fifth term from the end is the seventh from the beginning. The fifth term is

T_{5} = (4 10) (2^{1/3})^{6} (\frac{1}{2 \cdot 3 ^{1/3}})^{4} = (4 10) 2^{- 2} 3^{- 4/3}

, and the seventh is

T_{7} = (6 10) (2^{1/3})^{4} (\frac{1}{2 \cdot 3 ^{1/3}})^{6} = (4 10) 2^{- 14/3} 3^{- 2}

. Their ratio is

\frac{2 ^{- 2} 3 ^{- 4/3}}{2 ^{- 14/3} 3 ^{- 2}} = 2^{8/3} 3^{2/3} = (2^{8} \cdot 3^{2})^{1/3} = (2304)^{1/3} = 4 (36)^{1/3}

, i.e.

4 (36)^{1/3} : 1

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option