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JEE Main 2019 · 10 Jan · Shift I

JEE Main 2019 Mathematics question on Permutations and Combinations

If ∑i=120(20Ci−120Ci+20Ci−1)3=k21\sum_{i=1}^{20} \left( \frac{^{20}C_{i-1}}{^{20}C_i + ^{20}C_{i-1}} \right)^3 = \frac{k}{21}, then kk equals
  1. A.
    100
  2. B.
    400
  3. C.
    200
  4. D.
    50
Show answer and solution

Correct option: A

Solution

The key simplification: ²⁰Ci-1/(²⁰Ci + ²⁰Ci-1) = ²⁰Ci-1/²¹Ci (by Pascal's identity ²⁰Ci + ²⁰Ci-1 = ²¹Ci).

Now ²⁰Ci-1/²¹Ci = [20!/(i-1)!(21-i)!] / [21!/(i!(21-i)!)] = i/21.

So the sum becomes Σ(i=1 to 20) (i/21)³ = (1/21³) · Σ(i=1 to 20) i³ = (1/21³) · [20·21/2]² = (1/21³) · (210)² = (44100)/(9261) = 100/21.

Therefore k/21 = 100/21, giving k = 100.

[JEE Main 2019 · 10 Jan · Shift I]

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