JEE Main 2023 · 29 January · evening · Single correct · +4 / −1

Let K be the sum of the coefficients of the odd powers of x in the expansion of (1+x)⁹⁹. Let a be the middle term in the expansion of ( 2 + 1/√2 )²⁰⁰. If ²⁰⁰C₉₉K/a = 2ˡm/n, where m and n are odd numbers, then the ordered pair (l,n) is equal to

  1. (a) (50, 101)
  2. (b) (50, 51)
  3. (c) (51, 101)
  4. (d) (51, 99)
Answer and solution

Answer (a) (50, 101)

Solution

Maths sir · 4 steps

Find K and the middle term separately, then the binomial coefficients cancel almost completely.

  1. Find K. Put x = 1 and x = -1 into (1+x)⁹⁹ and subtract. The even powers cancel, so the odd-power coefficients sum to K = (2⁹⁹ - 0)/2 = 2⁹⁸.
  2. Middle term. With power 200 there are 201 terms, so the middle one is T₁₀₁. It is a = ²⁰⁰C₁₀₀ 2¹⁰⁰ (1/√2)¹⁰⁰ = ²⁰⁰C₁₀₀ 2¹⁰⁰ × 2⁻⁵⁰ = ²⁰⁰C₁₀₀ 2⁵⁰.

    (²⁰⁰C₉₉ K)/a = ²⁰⁰C₉₉/²⁰⁰C₁₀₀ × 2⁹⁸/2⁵⁰ = 100/101 × 2⁴⁸

  3. Ratio of coefficients. ²⁰⁰C₉₉/²⁰⁰C₁₀₀ = (100! 100!)/(99! 101!) = 100/101. Now 100 = 2² × 25, so the value is (2⁵⁰ × 25)/101.
  4. So. l = 50, m = 25 and n = 101, with m and n both odd. The ordered pair is (50, 101), option (a).

The idea to keep. Sum of odd-power (or even-power) coefficients of (1+x)ⁿ is 2ⁿ⁻¹. For ⁿCᵣ/ⁿCᵣ₊₁, use (r+1)/(n-r) instead of expanding factorials.

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.