# Show that the Expansion of ( X 2 + 1 X ) 12 Does Not Contain Any Term Involving X−1. - Mathematics

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

#### Solution

Suppose x1 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.

Concept: Binomial Theorem - Simple Applications of Binomial Theorem
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 12 | Page 38