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NDA 1 2024 Maths, question 34

Topic: Algebra

Q34

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

  1. (a) 11Correct answer
  2. (b) 22
  3. (c) 44
  4. (d) 88

Solution.

Answer: (a) 11

  1. (x+y)2n+1(x−y)2n+1=(x2−y2)2n+1(x + y)^{2n+1}(x - y)^{2n+1} = (x^{2} - y^{2})^{2n+1}, which has two middle terms.
  2. Their sum is (−1)n(2n+1n)x2ny2n(x2−y2)(-1)^{n}\binom{2n+1}{n}x^{2n}y^{2n}(x^{2} - y^{2}).
  3. For nonzero x,yx, y this is zero only when x2=y2x^{2} = y^{2}, so x2y2=1\frac{x^{2}}{y^{2}} = 1.

Trap: x=0x = 0 also makes the middle terms zero, but then the ratio is 0; the question means nonzero xx and yy.