Question

The last term in the binomial expansion of ${\left(\sqrt[3]{2}-\frac{1}{\sqrt{2}}\right)}^n$ is ${\left(\frac{1}{3\sqrt[3]{9}}\right)}^{lo{g}_3^8}$ then the 5th term from the beginning is

• A. ${}^{10}{c}_6$
• B. $2{}^{.10}{c}_4$
• C. $\frac{1}{2}.10c_4$
• D. none of these.

Answer 1

Answer: (A)

The last term $={}^n{c}_n\cdot {\left(-\frac{1}{\sqrt{2}}\right)}^n={\left(\frac{1}{3\sqrt[3]{9}}\right)}^{log{3}^8}$
(As given in the problem)
$\therefore {\left(-1\right)}^n{\left(\frac{1}{2}\right)}^{n2}={\left(\frac{1}{3^{53}}\right)}^{log{3}^8}$
$=3^{-\frac{5}{3}\cdot 3lo{g}_{3^2}}$
$=3^{-5lo{g}_{3^2}}=3^{lo{g}_3{2}^{-5}}$
$=2^{-5}={\left(\frac{1}{2}\right)}^5$
$\therefore n=10$
$\therefore T_5=T_{4+1}={}^{10}{C}_4{\left(\sqrt[2]{3}\right)}^6{\left(\frac{-1}{\sqrt{2}}\right)}^4$
$={}^{10}{C}_4{\left(2\right)}^2\frac{1}{2^2}$
$={}^{10}{C}_4={}^{10}{C}_6$.