JEE Main 2021 · 18 March · morning · Single correct · +4 / −1

Let (1 + x + 2x²)²⁰ = a₀ + a₁x + a₂x² + .... + a₄₀x⁴⁰. Then a₁ + a₃ + a₅ + ..... + a₃₇ is equal to

  1. (a) 2²⁰(2²⁰ - 21)
  2. (b) 2¹⁹(2²⁰ - 21)
  3. (c) 2¹⁹(2²⁰ + 21)
  4. (d) 2²⁰(2²⁰ + 21)
Answer and solution

Answer (b) 2¹⁹(2²⁰ - 21)

Solution

Maths sir · 4 steps

Put x = 1 and x = -1 to get all the odd coefficients, then remove the one extra term, a₃₉.

  1. Odd coefficients. Let f(x) = (1 + x + 2x²)²⁰. Then f(1) - f(-1) cancels the even-index terms and doubles the odd-index ones. So a₁ + a₃ + … + a₃₉ = (f(1) - f(-1))/2.

    a₁ + a₃ + … + a₃₉ = (4²⁰ - 2²⁰)/2 = (2⁴⁰ - 2²⁰)/2 = 2¹⁹(2²⁰ - 1)

  2. One term too many. The question stops at a₃₇, so subtract a₃₉. To reach x³⁹ from 20 factors, the only way is to take 2x² from 19 of them and x from the remaining one, since 2 × 19 + 1 = 39.
  3. Find a39. The single x can come from any of the 20 factors, so a₃₉ = 20 × 2¹⁹.
  4. So. a₁ + a₃ + … + a₃₇ = 2¹⁹(2²⁰ - 1) - 20 × 2¹⁹ = 2¹⁹(2²⁰ - 21), which is option (b).

The idea to keep. For sums of odd or even coefficients, use (f(1) ∓ f(-1))/2. If the sum stops early, find the missing end terms from the highest powers and subtract them.

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 2021.