JEE Main 2019 — Mathematics Algebra

Total number of terms in the expansion of ${({2}^{\frac{1}{3}}+\frac{1}{2{(3)}^{\frac{1}{3}}})}^{10}=10+1=11$(231​+2(3)31​1​)10=10+1=11

Now, ${5}^{th}$5th term from end $=11-5+1={7}^{th}$=11−5+1=7th from beginning.

We need to find a ratio of the ${5}^{th}$5th term from the beginning to the ${5}^{th}$5th term from the end (i.e. ${7}^{th}$7th term from beginning) in the binomial expansion of ${({2}^{\frac{1}{3}}+\frac{1}{2{(3)}^{\frac{1}{3}}})}^{10}=\frac{{T}_{5}}{{T}_{7}}$(231​+2(3)31​1​)10=T7​T5​​

$$=\frac{{}^{10}{C}_{4}{({2}^{\frac{1}{3}})}^{10-4}{(\frac{1}{2{(3)}^{\frac{1}{3}}})}^{4}}{{}^{10}{C}_{6}{({2}^{\frac{1}{3}})}^{10-6}{(\frac{1}{2{(3)}^{\frac{1}{3}}})}^{6}}[\because {T}_{r+1}{=}^{n}{C}_{r}{(a)}^{n-r}{(b)}^{r}]$$=10C6​(231​)10−6(2(3)31​1​)610C4​(231​)10−4(2(3)31​1​)4​[∵Tr+1​=nCr​(a)n−r(b)r]

$$=\frac{\frac{(10!)}{(4!)(6!)}\times {(2)}^{2}\times \frac{1}{({2}^{4}){(3)}^{\frac{4}{3}}}}{\frac{(10!)}{(6!)(4!)}\times {(2)}^{\frac{4}{3}}\times \frac{1}{({2}^{6}){(3)}^{2}}}$$=(6!)(4!)(10!)​×(2)34​×(26)(3)21​(4!)(6!)(10!)​×(2)2×(24)(3)34​1​​

$$=4{(36)}^{\frac{1}{3}}:1$$=4(36)31​:1