JEE Main 2023 · 11 April · evening · Single correct · +4 / −1

The sum of the coefficients of three consecutive terms in the binomial expansion of (1+x)ⁿ⁺², which are in the ratio 1: 3: 5, is equal to :

  1. (a) 63
  2. (b) 92
  3. (c) 25
  4. (d) 41
Answer and solution

Answer (a) 63

Solution

Maths sir · 4 steps

Divide neighbouring coefficients: each ratio gives one clean linear equation in N and r.

  1. Set up. Write N = n + 2. Let the three coefficients be C(N, r-1), C(N, r), C(N, r+1). The useful fact is C(N, r)/C(N, r-1) = (N-r+1)/r.

    (N-r+1)/r = 3 and (N-r)/(r+1) = 5/3

  2. Solve the pair. The first gives N = 4r - 1. The second gives 3N = 8r + 5. Substituting, 12r - 3 = 8r + 5, so r = 2 and N = 7.
  3. Check the coefficients. They are C(7, 1) = 7, C(7, 2) = 21, C(7, 3) = 35. The ratio 7 : 21 : 35 is indeed 1 : 3 : 5.
  4. So. The sum is 7 + 21 + 35 = 63, which is option (a).

The idea to keep. For consecutive binomial coefficients, never expand factorials. Use C(N, r) ÷ C(N, r-1) = (N-r+1)/r, so each given ratio becomes a linear equation.

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.