Question

Show that the expansion of \[\left( x^2 + \frac{1}{x} \right)^{12}\]  does not contain any term involving x⁻¹.

 

 

Answer 1

Suppose x⁻¹ occurs at the (r + 1)th term in the given expression.
Then 

\[T_{r + 1} = ^{12}{}{C}_r ( x^2 )^{12 - r} \left( \frac{1}{x} \right)^r \]
\[ = ^ {12}{}{C}_r x^{24 - 2r - r} \]
\[\text{ For this term to contain } x^{- 1} , \text{ we must have } \]
\[24 - 3r = - 1\]
\[ \Rightarrow 3r = 25\]
\[ \Rightarrow r = \frac{25}{3}\]
\[\text{ It is not possible, as r is not an integer }  . \]

Hence, the expansion of \[\left( x^2 + \frac{1}{x} \right)^{12}\]  does not contain any term involving x⁻¹.