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Find the middle terms(s) in the expansion of: (v) ( x − 1 x ) 2 n + 1

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प्रश्न

Find the middle terms(s) in the expansion of:

(v) \[\left( x - \frac{1}{x} \right)^{2n + 1}\]

 

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उत्तर

\[\left( x - \frac{1}{x} \right)^{2n + 1} \]
\[\text{ Here, } \left( 2n + 1 \right) \text{ is an odd number .}  \]
\[\text{ Therefore, the middle terms are } \left( \frac{2n + 1 + 1}{2} \right)\text{ th and} \left( \frac{2n + 1 + 1}{2} + 1 \right)\text{ th i . e . (n + 1)th and (n + 2)th terms }  . \]
\[\text{ Now, we have: }  \]
\[ T_{n + 1} \]
\[ =^{2n + 1}{}{C}_n x^{2n + 1 - n} \times \frac{( - 1 )^n}{x^n}\]
`=(-1)^n  "^(2n+1)C_n x`
\[\text{ And,}  \]
\[ T_{n + 2} = T_{n + 1 + 1} \]
\[ =^{2n + 1}{}{C}_{n + 1} x^{2n + 1 - n - 1} \frac{( - 1 )^{n + 1}}{x^{n + 1}}\]
\[ = ( - 1 )^{n + 1} \]  ` "^{2n + 1}C_{n + 1} \times \frac{1}{x}`

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अध्याय 18: Binomial Theorem - Exercise 18.2 [पृष्ठ ३८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.2 | Q 15.05 | पृष्ठ ३८

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