Question

The last term in the binomial expansion of $\left(\sqrt[3]{2}-\frac{1}{\sqrt{2}}\right)^n$ is $\left(\frac{1}{3\sqrt[3]{9}}\right)^{log_3^8}$ then the 5th term from the beginning is

• A. $\,{}^{10}c_6$
• B. $2\,{}^{.10}c_4$
• C. $\frac{1}{2}\,{.10}c_4$
• 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)
$\therefore \left(-1\right)^{n} \left(\frac{1}{2}\right)^{n 2} = \left(\frac{1}{3^{5 3}}\right)^{log\,3^8}$
$= 3^{-\frac{5}{3}\cdot3\,log_{3^{2}}}$
$=3^{-5\,log_{3^2}}=3^{log_{3}\,2^{-5}}$
$= 2^{-5}=\left(\frac{1}{2}\right)^{5}$
$\therefore n = 10$
$\therefore T_{5} = T_{4+1}=\,{}^{10}C_{4}\left(\sqrt[2]{3}\right)^{6} \left(\frac{-1}{\sqrt{2}}\right)^{4}$
$= \,{}^{10}C_{4}\left(2\right)^{2} \frac{1}{2^{2}}$
$= \,{}^{10}C_{4} = \,{}^{10}C_{6}$.