JEE Main 2024 · 1 February · evening · Single correct · +4 / −1

Let m and n be the coefficients of seventh and thirteenth terms respectively
in the expansion of (1/3 x^(1/3)+1/(2 x^(2/3)))¹⁸. Then (n/m)^(1/3) is :

  1. (a) 1/9
  2. (b) 1/4
  3. (c) 4/9
  4. (d) 9/4
Answer and solution

Answer (d) 9/4

Solution

Maths sir · 4 steps

Divide n by m: the binomial coefficients cancel, leaving only powers of 3 and 2.

  1. General term. In (a+b)¹⁸ the term Tᵣ₊₁ is C(18, r)a¹⁸⁻ʳbʳ. Here a = 1/3x^(1/3) and b = 1/2x^(-2/3). The coefficient is the part without x.

    coeff of Tᵣ₊₁ = C(18, r)(1/3)¹⁸⁻ʳ(1/2)ʳ

  2. Pick the terms. The seventh term has r = 6, so m = C(18, 6)(1/3)¹²(1/2)⁶. The thirteenth term has r = 12, so n = C(18, 12)(1/3)⁶(1/2)¹².
  3. Divide. C(18, 12) = C(18, 6), so the binomial coefficients cancel. That leaves n/m = 3¹²/3⁶ × 2⁶/2¹² = 3⁶/2⁶ = (3/2)⁶.
  4. So. (n/m)^(1/3) = (3/2)² = 9/4, which is option (d).

The idea to keep. The kth term has r = k - 1. For terms with r and 18 - r, C(18, r) = C(18, 18-r), so their ratio keeps only the powers of the two parts.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2024.