Question

The last term in $(2^{1/3}+2^{-1/2}{)}^{n}is{\left(\frac{1}{3(9{)}^{1/3}}\right)}^{lo{g}_{3}3}$ then show that the fifth term is 210

Answer 1

The last term = $T_{n+1}$
$={}^n{C}_n\left(2^{1/3}{)}^0\right(2^{-1/2}{)}^n={\left[\frac{1}{3(9{)}^{1/3}}\right]}^{lo{g}_{3}8}$
or $2^{-n/2}=(3^{-5/3}{)}^{lo{g}_{3}8}=3^{lo{g}_3{8}^{-5/3}}$
$\because a^{lo{g}_{a}n}=n$
or $2^{-n/2}=8^{-5/3}=(2^3{)}^{-5/3}=2^{-5}\therefore n=10$
$\therefore T_5={}^{10}{C}_4\left(2^{1/3}{)}^6\right(2^{-1/2}{)}^4$
= 210 $\left(2^2\right)\left(2^{-2}\right)$ = 210