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NDA 2012
Exam
NDA
Year
2012
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2012 · Mathematics · Binomial Theorem

The value of the term independent of x in the expansion of...

The value of the term independent of xx in the expansion of (x2−1x)9\left(x^2-\frac{1}{x}\right)^9 is

Options

  1. A9
  2. B18
  3. C48
  4. D84

Correct answer

D. 84

Step-by-step solution

(x2−1x)9tr+1=9Cr(x2)9−r(−1x)r=9Cr(x)18−2r(−1)rx−r\begin{aligned} \left(x^2-\frac{1}{x}\right)^9 & \\ t_{r+1} & ={ }^9 C_r\left(x^2\right)^{9-r}\left(\frac{-1}{x}\right)^r \\ & ={ }^9 C_r(x)^{18-2 r}{ }_{(-1)^r x^{-r}} \end{aligned} Put 18−3r=0r=618-3 r=0r=6 Put r=6r=6, in (1), t7=9C6(1)t_7={ }^9 C_6(1)^6=9!6!3!=9⋅8⋅7⋅6!6!⋅3⋅2=84.\frac{9!}{6!3!}=\frac{9 \cdot 8 \cdot 7 \cdot 6!}{6!\cdot 3 \cdot 2}=84 .
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