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If the expansion of (x-1x2)2n contains a term independent of x, then n is a multiple of 2. - Mathematics

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Question

If the expansion of `(x - 1/x^2)^(2n)` contains a term independent of x, then n is a multiple of 2.

Options

  • True

  • False

MCQ
True or False
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Solution

This statement is False.

Explanation:

The given expression is `(x - 1/x^2)^(2n)` 

Tr+1 = `""^(2n)C_r (x)^(2n - r) (- 1/x^2)^r`

= `""^(2n)C_r (x)^(2n - r) (-1)^r * 1/x^(2r)`

= `""^(2n)C_r (x)^(2n - r - 2r) (-1)^r`

= `""^(2n)C_r (x)^(2n - 3r) (-1)^r`

For the term independent of x, 2n – 3r = 0

∴ r = `(2n)/3` which not an integer and the expression is not possible to be true

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Chapter 8: Binomial Theorem - Exercise [Page 146]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 8 Binomial Theorem
Exercise | Q 39 | Page 146

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