JEE Main 2023 · 24 January · evening · Single correct · +4 / −1

If (³⁰C₁)² + 2(³⁰C₂)² + 3(³⁰C₃)² + ... + 30(³⁰C₃₀)² = (α 60!)/(30!)² then α is equal to :

  1. (a) 30
  2. (b) 10
  3. (c) 15
  4. (d) 60
Answer and solution

Answer (c) 15

Solution

Maths sir · 3 steps

Pair the sum with itself read backwards, and the awkward factor r turns into a constant 30.

  1. Reverse the sum. Call the sum S = Σᵣ₌₀³⁰ r(³⁰Cᵣ)². The r = 0 term is zero, so adding it changes nothing. Since ³⁰Cᵣ = ³⁰C₃₀₋ᵣ, replacing r by 30 - r gives S = Σᵣ₌₀³⁰ (30 - r)(³⁰Cᵣ)².
  2. Add the two forms. Adding the two expressions makes r + (30 - r) = 30 in every term. So 2S = 30Σᵣ₌₀³⁰(³⁰Cᵣ)².

    Σᵣ₌₀ⁿ(ⁿCᵣ)² = ²ⁿCₙ ⇒ 2S = 30 × ⁶⁰C₃₀

  3. So. S = 15 × ⁶⁰C₃₀ = (15 × 60!)/(30!)². Therefore α = 15, option (c).

The idea to keep. When a binomial sum carries a weight r, add the sum to itself with r replaced by n - r. The weights add up to n and you are left with a standard identity.

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.