Constant term in the expansion of \[\left( x - \frac{1}{x} \right)^{10}\] is

−252Suppose (r + 1)th term is the constant term in the given expansion.Then, we have: \[T_{r + 1} = ^{10}{}{C}_r (x )^{10 - r} \left( \frac{- 1}{x} \right)^r \]\[ = ^{10}{}{C}_r ( - 1 )^r x^{10 - r - r} \]\[\text{ For this term to be constant, we must have: } \]\[10 - 2r = 0\]\[ \Rightarrow r = 5\]\[ \therefore \text{ Required term } = - ^{10}{}{C}_5 = - 252\]

Constant Term in the Expansion of ( X − 1 X ) 10 Is(A) 152 (B) −152 (C) −252 (D) 252 - Mathematics

Question

Constant term in the expansion of ${(x - \frac{1}{x})}^{10}$ is

Options

152
−152
−252
252

MCQ

Solution

−252

Suppose (r + 1)th term is the constant term in the given expansion.
Then, we have:

$T_{r + 1} =^{10} C_{r} (x)^{10 - r} {(\frac{- 1}{x})}^{r}$
$=^{10} C_{r} (- 1)^{r} x^{10 - r - r}$
$For this term to be constant, we must have:$
$10 - 2 r = 0$
$\Rightarrow r = 5$
$∴ Required term = -^{10} C_{5} = - 252$

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Q 32 Q 31 Q 33

Chapter 18: Binomial Theorem - Exercise 18.4 [Page 48]

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Chapter 18 Binomial Theorem
Exercise 18.4 | Q 32 | Page 48

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Constant Term in the Expansion of ( X − 1 X ) 10 Is(A) 152 (B) −152 (C) −252 (D) 252 - Mathematics

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Constant Term in the Expansion of ( X − 1 X ) 10 Is(A) 152 (B) −152 (C) −252 (D) 252 - Mathematics

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