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 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
$t_{n+1}={{}^{n}C}_n{\left(2^{\frac{1}{3}}\right)}^{n-n}{\left(\frac{1}{\sqrt{2}}\right)}^n=\frac{{(-1)}^n}{2^{\frac{n}{2}}}$

Also, we have

${\left(\frac{1}{3^{\frac{5}{3}}}\right)}^{{\log }_{3}8}=\frac{1}{{\left(3^{\frac{5}{3}}\right)}^{3{\log }_{3}2}}=3^{-\left(\frac{5}{3}\right){\log }_3{2}^3}=2^{-5}$

Thus,

$\frac{{(-1)}^n}{2^{\frac{n}{2}}}=2^{-5}\Rightarrow \frac{{(-1)}^n}{2^{\frac{n}{2}}}=\frac{{(-1)}^n}{2^5}$

$\Rightarrow \frac{n}{2}=5\Rightarrow n=10$

Now, $t_5=t_{4+1}={{}^{10}C}_4{\left(2^{\frac{1}{3}}\right)}^{10-4}{(-\frac{1}{\sqrt{2}})}^4$

$=\frac{10!}{4!6!}{\left(2^{\frac{1}{3}}\right)}^6{(-1)}^4{\left(2^{-\frac{1}{2}}\right)}^4=210$