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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Binomial Theorem

NDA 2024 · Mathematics · Binomial Theorem

In a binomial expansion of (x+y)^ 2 n+1 (x-y)^ 2 n+1 , the sum of...

In a binomial expansion of (x+y)2n+1(x−y)2n+1(x+y)^{2 n+1}(x-y)^{2 n+1}, the sum of middle terms is zero. What is the value of (x2y2)\left(\frac{x^2}{y^2}\right) ?

Options

  1. A1
  2. B2
  3. C4
  4. D8

Correct answer

A. 1

Step-by-step solution

(x+y)2n+1⋅(x−y)2n+1(x+y)^{2 n+1} \cdot(x-y)^{2 n+1} Middle term, 2n+22,2n+42=(n+1) th term, (n+2) th term Tn+1=2n+1Cn(x2)n+1(y2)nTn+2=2n+1Cn+1(x2)n(y2)n+12n+1Cn(x2)n+1(y2)n=2n+1Cn+1(x2)n(y2)n+1⇒2n+1Cn2n+1Cn+1=x2ny2n+2x2n+2y2n⇒(2n+1)!(n+1)!n!×n!(n+1)!(2n+1)!=y2x2⇒y2x2=1∵1y2=11∴x2y2=1\begin{aligned} & \frac{2 n+2}{2}, \frac{2 n+4}{2}=(n+1) \text{ th term, }(n+2) \text{ th term } \\ & \mathrm{T}_{n+1}={ }^{2 n+1} \mathrm{C}_n\left(x^2\right)^{n+1}\left(y^2\right)^n \\ & \mathrm{ T}_{n+2}=2^{n+1} \mathrm{C}_{n+1}\left(x^2\right)^n\left(y^2\right)^{n+1} \\ & { }^{2 n+1} \mathrm{C}_n\left(x^2\right)^{n+1}\left(y^2\right)^n=2^{n+1} \mathrm{C}_{n+1}\left(x^2\right)^n\left(y^2\right)^{n+1} \\ & \Rightarrow \quad \frac{2^{n+1} \mathrm{C}_n}{{ }^{2 n+1} \mathrm{C}_{n+1}}=\frac{x^{2 n} y^{2 n+2}}{x^{2 n+2} y^{2 n}} \\ & \Rightarrow \frac{(2 n+1)!}{(n+1)!n!} \times \frac{n!(n+1)!}{(2 n+1)!}=\frac{y^2}{x^2} \\ & \Rightarrow \quad \frac{y^2}{x^2}=1 \\ & \because \quad \frac{1}{y^2}=\frac{1}{1} \\ & \therefore \quad \frac{x^2}{y^2}=1 \end{aligned}
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