JEE Main 2021 · 24 February · evening · Single correct · +4 / −1

If n ≥ 2 is a positive integer, then the sum of the series ⁿ⁺¹C₂ + 2( ²C₂ + ³C₂ + ⁴C₂ + ... + ⁿC₂ ) is :

  1. (a) n(2n + 1)(3n + 1)/6
  2. (b) n(n + 1)(2n + 1)/6
  3. (c) n(n + 1)²(n + 2)/12
  4. (d) n(n - 1)(2n + 1)/6
Answer and solution

Answer (b) n(n + 1)(2n + 1)/6

Solution

Maths sir · 4 steps

Use the hockey-stick identity to collapse the bracket into one binomial coefficient, then factor.

  1. Hockey-stick identity. Adding ʳC₂ from r = 2 up to r = n gives one coefficient one level higher: ²C₂ + ³C₂ + … + ⁿC₂ = ⁿ⁺¹C₃.

    S = ⁿ⁺¹C₂ + 2 ⁿ⁺¹C₃ = (n+1)n/2 + 2(n+1)n(n-1)/6

  2. Factor out. Take n(n+1)/6 out of both terms. This leaves S = n(n+1)/6[3 + 2(n-1)] = n(n+1)(2n+1)/6.
  3. Quick check. At n = 2 the series is ³C₂ + 2(²C₂) = 3 + 2 = 5. Option (b) gives (2 × 3 × 5)/6 = 5, which matches. It is also 1² + 2², the sum of the first n squares.
  4. So. The sum is n(n+1)(2n+1)/6, which is option (b).

The idea to keep. A running sum of ʳCₖ with the bottom index fixed always collapses to ⁿ⁺¹Cₖ₊₁. Then test n = 2 against the options to confirm in seconds.

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.