Question

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

• A. 210
• B. 420
• C. 105
• D. 425

Answer 1

Answer: (A)

Last term $={}^n{C}_n\cdot {\left(2^{\frac{1}{3}}\right)}^0\cdot {\left(-\frac{1}{\sqrt{2}}\right)}^n=\frac{(-1{)}^n}{2^{n/2}}={\left(\frac{1}{3^{5/3}}\right)}^{{\log }_{2}8}=3^{-\frac{5}{3}\cdot 3}=2^{-5}$
$\Rightarrow \frac{n}{2}=-5\Rightarrow n=10$
$t_5={}^{10}{C}_4{\left(2^{\frac{1}{3}}\right)}^{10-4}{\left(-\frac{1}{\sqrt{2}}\right)}^4=210$