Question

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

Last term of ${(2^{\frac{1}{3}}-\frac{1}{\sqrt{2}})}^n$ is
$t_{n+1}{=}^n{C}_n{\left(2^{\frac{1}{3}}\right)}^{n-n}{(-\frac{1}{\sqrt{2}})}^n{=}^n{C}_n(-1{)}^n\frac{1}{2^{n/2}}=\frac{(-1{)}^n}{2^{n/2}}$
Also, we have
${\left(\frac{1}{3^{5/3}}\right)}^{lo{g}_{3}8}=\frac{1}{{\left(3^{5/3}\right)}^{3lo{g}_{3}2}}=3^{-\left(\frac{5}{3}lo{g}_3{2}^3\right)}=2^{-5}Thus,\frac{(-1{)}^n}{2^{n/2}}=2^{-5}\Rightarrow \frac{(-1{)}^n}{2^{n/2}}=\frac{(-1{)}^{10}}{2^5}\Rightarrow \frac{n}{2}=5\Rightarrow n=10Now,t_5=t_{4+1}{=}^{10}{C}_4{\left(2^{\frac{1}{3}}\right)}^{10-4}{(-\frac{1}{\sqrt{2}})}^4=\frac{10!}{4!6!}{\left(2^{\frac{1}{3}}\right)}^6(-1{)}^4{(2-\frac{1}{2})}^4=210(2^2\left)\right(1\left)\right(2^{-2})=210$