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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2014 · Mathematics · Binomial Theorem

Paragraph: In the expansion of (x^3-(1)/(x^2))^n , where n is a...

Paragraph: In the expansion of (x3−1x2)n\left(x^3-\frac{1}{x^2}\right)^n, where nn is a positive integer, the sum of the coefficients of x5andx10x^5 and x^{10} is 0.
Question: What is nn equal to?

Options

  1. A5
  2. B10
  3. C15
  4. DNone of the above

Correct answer

C. 15

Step-by-step solution

(x3−1x2)n=nC0(x3)n−nC1(x3)n−1⋅(1x2)+nC2(x3)n−2⋅(1x2)2−nC3(x3)n−3(1x2)3+nC4(x3)n−4(1x2)4−nC5(x3)n−5(1x2)5+nC6(x3)n−6(1x2)6−nC7(x3)n−7⋅(1x2)7+nC8(x3)n−8(1x2)8−nC9(x3)n−9⋅(1x2)9\begin{aligned} & \left(x^3-\frac{1}{x^2}\right)^n \\ & ={ }^n \mathrm{C}_0\left(x^3\right)^n-{ }^n \mathrm{C}_1\left(x^3\right)^{n-1} \cdot\left(\frac{1}{x^2}\right)+{ }^n \mathrm{C}_2\left(x^3\right)^{n-2} \cdot\left(\frac{1}{x^2}\right)^2 \\ & -{ }^n \mathrm{C}_3\left(x^3\right)^{n-3}\left(\frac{1}{x^2}\right)^3+{ }^n \mathrm{C}_4\left(x^3\right)^{n-4}\left(\frac{1}{x^2}\right)^4-{ }^n \mathrm{C}_5\left(x^3\right)^{n-5} \\ & \left(\frac{1}{x^2}\right)^5+{ }^n \mathrm{C}_6\left(x^3\right)^{n-6}\left(\frac{1}{x^2}\right)^6-{ }^n \mathrm{C}_7\left(x^3\right)^{n-7} \cdot\left(\frac{1}{x^2}\right)^7 \\ & +{ }^n \mathrm{C}_8\left(x^3\right)^{n-8}\left(\frac{1}{x^2}\right)^8-{ }^n \mathrm{C}_9\left(x^3\right)^{n-9} \cdot\left(\frac{1}{x^2}\right)^9 \end{aligned}⇒nC8=nC7⇒n=8+7⇒n=15\begin{aligned} & \Rightarrow{ }^n \mathrm{C}_8={ }^n \mathrm{C}_7 \Rightarrow n=8+7 \Rightarrow n=15 \end{aligned}Hence,Hence,n=15$.
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