Question

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

• A. $420$
• B. $210$
• C. $105$
• D. $84$

Answer 1

The correct option is B $210$

Last term of ${(2^{1/3}-\frac{1}{\sqrt{2}})}^n$ is $T_{n+1}={}^n{C}_n(2^{1/3}{)}^{n-n}{(-\frac{1}{\sqrt{2}})}^n={}^n{C}_n(-1{)}^n\frac{1}{2^{n/2}}=\frac{(-1{)}^n}{2^{n/2}}$
Also, it is given that last term is
${\left(\frac{1}{3^{5/3}}\right)}^{{\log }_{3}8}=3^{-\left(5/3\right){\log }_3{2}^3}=2^{-5}$
Thus, $\frac{(-1{)}^n}{2^{n/2}}=2^{-5}$
$\Rightarrow \frac{(-1{)}^n}{2^{n/2}}=\frac{1}{2^5}$

$\Rightarrow \frac{n}{2}=5$
$\Rightarrow n=10$
Now, $T_5=T_{4+1}={}^{10}{C}_4(2^{1/3}{)}^{10-4}{(-\frac{1}{\sqrt{2}})}^4$
$=\frac{10!}{4!6!}\left(2^{1/3}{)}^6\right(-1{)}^4(2^{-1/2}{)}^4$
$=210\left(2^2\right)\left(1\right)\left(2^{-2}\right)=210$