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 $={ }^n C_n \cdot\left(2^{\frac{1}{3}}\right)^0 \cdot\left(-\frac{1}{\sqrt{2}}\right)^n=\frac{(-1)^n}{2^{n / 2}}=\left(\frac{1}{3^{5 / 3}}\right)^{\log _2 8}=3^{-\frac{5}{3} \cdot 3}=2^{-5}$
$\Rightarrow \frac{n}{2}=-5 \Rightarrow n=10$
$t_5={ }^{10} C_4\left(2^{\frac{1}{3}}\right)^{10-4}\left(-\frac{1}{\sqrt{2}}\right)^4=210$