The sum of the series 2 × 1 × ²⁰ C₄-3 × 2 × ²⁰ C₅+4 × 3 × ²⁰ C₆-5 × 4 × ²⁰ C₇+… …+18 × 17 × ²⁰ C₂₀, is equal to ____________.

Solution

Maths sir · 4 steps

Extend the sum to start from r = 0: the full alternating sum is 0, so only a few missing terms are left.

1. General term. The term with ²⁰Cᵣ is (-1)ʳ (r-2)(r-3) ²⁰Cᵣ, for r = 4 to 20. Check: r = 4 gives +2 × 1, r = 5 gives -3 × 2, r = 20 gives +18 × 17.
2. Full sum is zero. Write (r-2)(r-3) = r(r-1) - 4r + 6. Each of Σ (-1)ʳ ²⁰Cᵣ, Σ (-1)ʳ r ²⁰Cᵣ = -20(1-1)¹⁹ and Σ (-1)ʳ r(r-1) ²⁰Cᵣ = 20 × 19 (1-1)¹⁸ is 0 when r runs from 0 to 20.

   Σᵣ₌₀²⁰ (-1)ʳ (r-2)(r-3) ²⁰Cᵣ = 0

3. Missing terms. The given series S lacks r = 0, 1, 2, 3. At r = 0 the term is (+1)(-2)(-3)(1) = 6. At r = 1 it is (-1)(-1)(-2)(20) = -40. At r = 2 and r = 3 the factor (r-2)(r-3) is 0.
4. So. 6 - 40 + S = 0, which gives S = 34.

The common slip: -34. Moving the missing terms 6 and -40 to the other side with the wrong sign gives -34. Write 6 - 40 + S = 0 before solving for S.

The idea to keep. For n large, Σᵣ₌₀ⁿ (-1)ʳ ⁿCᵣ P(r) = 0 whenever P is a polynomial of degree less than n. Complete the sum, then subtract the few missing terms.

Answer from the NTA final answer key. Solution by PrepWiser.