JEE Main 2023 · 15 April · morning · Single correct · +4 / −1

Let (a+b x+c x²)¹⁰=Σᵢ₌₀²⁰ pᵢ xⁱ, a, b, c ∈ ℕ.
If p₁=20 and p₂=210, then 2(a+b+c) is equal to :

  1. (a) 15
  2. (b) 8
  3. (c) 6
  4. (d) 12
Answer and solution

Answer (d) 12

Solution

Maths sir · 5 steps

Read off p₁ and p₂ by picking terms from the ten brackets, and natural numbers pin a down at once.

  1. Coefficient of x. To get x¹, take bx from one bracket and a from the other nine. There are C(10, 1) = 10 ways, so p₁ = 10a⁹ b = 20, giving a⁹ b = 2.
  2. Use natural numbers. With a, b ∈ ℕ, a ≥ 2 would make a⁹ ≥ 512, which is too big. So a = 1 and b = 2.
  3. Coefficient of x squared. There are two ways to get x²: take cx² from one bracket, or take bx from two brackets. All other brackets give a.

    p₂ = C(10, 1)a⁹ c + C(10, 2)a⁸ b² = 10c + 45 × 4 = 10c + 180

  4. Find c. 10c + 180 = 210 gives c = 3.
  5. So. a + b + c = 1 + 2 + 3 = 6, and 2(a + b + c) = 12, which is option (d).

The idea to keep. For low powers of x in (a + bx + cx²)ⁿ, count which brackets give which term. For x², never forget the bx · bx route alongside the single cx².

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.