Question

A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of

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

Answer 1

The correct option is B $4\left(36\right)\frac{1}{3}:1$

${[2^{\frac{1}{3}}+\frac{1}{2(3{)}^{\frac{1}{3}}}]}^{10}$

$\Rightarrow 5^{th}$ term of given sequence from beginning is

$a{=}^{10}{C}_4[2^{\frac{1}{3}}{]}^{10-4}{\left[\frac{1}{2(3{)}^{\frac{1}{3}}}\right]}^4$

$a{=}^{10}{C}_4(2^{\frac{1}{3}}{)}^6{\left[\frac{1}{2(3{)}^{\frac{1}{3}}}\right]}^4$ -------- (1)

$5^{th}$ term of given sequence from last is

$b{=}^{10}{C}_6[2^{\frac{1}{3}}{]}^{10-6}{\left[\frac{1}{2(3{)}^{\frac{1}{3}}}\right]}^6$ ------

$b{=}^{10}{C}_6(2{)}^{\frac{4}{3}}\frac{1}{\left(2{)}^6\right(3{)}^{\frac{6}{3}}}$ -------- (2)

Now

$\frac{a}{b}=\frac{{}^{10}{C}_4(2{)}^{\frac{6}{3}}\frac{1}{\left(2{)}^4\right(3{)}^{\frac{4}{3}}}}{{}^{10}{C}_6(2{)}^{\frac{4}{3}}\frac{1}{\left(2{)}^6\right(3{)}^{\frac{6}{3}}}}{[}^{10}{C}_4{=}^{10}{C}_6]$

$\frac{a}{b}=\left(2{)}^{\frac{2}{3}}\right(2{)}^2.(3{)}^{\frac{2}{3}}$

$\frac{a}{b}=(2{)}^2.(4{)}^{\frac{1}{4}}.(9{)}^{\frac{1}{3}}$

$\left(\frac{a}{b}\right)=(2{)}^2.(36{)}^{\frac{1}{3}}$

$\frac{a}{b}=\frac{4(36{)}^{\frac{1}{3}}}{1}$

Hence $a:b=4(36{)}^{\frac{1}{3}}:1$.