Question

If the last term in the binomial expansion of $(2{}^{\frac{1}{3}}-\frac{1}{\sqrt{2}}){}^{n}is\left(\frac{1}{3^{\frac{5}{3}}}\right){}^{log{}_{3}8}$ , then the 5th term from the beginning is

• A. 210
• B. 420
• C. 105
• D. none of these

Answer 1

The correct option is A 210

Given,

${(2^{\frac{1}{3}}-\frac{1}{\sqrt{2}})}^n$

last term of $(x-a{)}^n=(-1{)}^n(a{)}^n$

last term of given equation,

$(-1{)}^n\left(\frac{1}{2^{\frac{n}{2}}}\right)={\left(\frac{1}{3\cdot 9^{\frac{1}{3}}}\right)}^{{\log }_{3}8}$

$={\left(\frac{1}{3\cdot 3^{\frac{2}{3}}}\right)}^{{\log }_{3}8}$

$={\left(\frac{1}{3^{\frac{5}{3}}}\right)}^{{\log }_{3}8}$

$=\left(\frac{1}{3^{\frac{5}{3}{\log }_{3}8}}\right)$

$=\left(\frac{1}{3^{{\log }_3{8}^{\frac{5}{3}}}}\right)$

$=\frac{1}{8^{\frac{5}{3}}}$

$=\frac{1}{2^5}$

$\Rightarrow \frac{n}{2}=5$

$\therefore n=10$

now,

${(2^{\frac{1}{3}}-\frac{1}{\sqrt{2}})}^{10}$

$T_5=T_{4+1}{=}^{10}{C}_4\left(2^{\frac{1}{3}}{)}^{10-4}{\left(\frac{1}{2^{\frac{1}{2}}}\right)}^4\right(-1{)}^4$

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

${=}^{10}{C}_4\left(2^2\right)\left(\frac{1}{2^2}\right)$

${=}^{10}{C}_4$

$=210$