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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.

 
 
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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
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