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Question

Show that the expansion of $$\left( x^2 + \dfrac{1}{x} \right)^{12}$$ does not contain any term involving $$x^{-1}$$.


Solution

General term is 
$$ T_{r} = ^{n}C_{r}\,x^{r}\,a^{n-r} $$
in $$ (x+a)r_{n}(r\epsilon z) $$
$$ \therefore T_{r} = ^{12}C_{r}(x^{2})^{r}\left ( \dfrac{1}{x} \right )^{12-r} $$
$$ \Rightarrow T_{r} = ^{12}C_{r}\,x^{3r-12} $$
for $$ x^{-1}$$ term $$ 3r-12 = -1 $$
$$ \Rightarrow \boxed{r = \dfrac{11}{3}} $$ not possible

1115303_529248_ans_f0c69f76e77a466bb816d0e711ae9d55.jpg

Mathematics

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