JEE Main 2023 · 8 April · morning · Single correct · +4 / −1

If the coefficients of three consecutive terms in the expansion of (1+x)ⁿ are in the ratio 1: 5: 20, then the coefficient of the fourth term is

  1. (a) 3654
  2. (b) 1827
  3. (c) 5481
  4. (d) 2436
Answer and solution

Answer (a) 3654

Solution

Maths sir · 4 steps

Turn each ratio of neighbouring coefficients into an equation in n and r, then solve the pair.

  1. Name the terms. Let the three consecutive coefficients be C(n, r-1), C(n, r) and C(n, r+1). For any two neighbours there is a short formula: C(n, r)/C(n, r-1) = (n-r+1)/r.

    (n-r+1)/r = 5, (n-r)/(r+1) = 20/5 = 4

  2. Solve the pair. The first equation gives n - r + 1 = 5r, so n = 6r - 1. The second gives n - r = 4r + 4, so n = 5r + 4. Equating them, 6r - 1 = 5r + 4, so r = 5 and n = 29.
  3. Fourth term. The general term is Tₖ₊₁ = C(n, k)xᵏ, so the fourth term has coefficient C(29, 3) = (29 × 28 × 27)/6 = 29 × 14 × 9.
  4. So. The coefficient of the fourth term is 29 × 126 = 3654, option (a).

The idea to keep. For consecutive binomial coefficients, use C(n, r)/C(n, r-1) = (n-r+1)/r. Two ratios give two linear equations in n and r. Remember that Tₖ₊₁ carries C(n, k).

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 2023.