The sum, of the coefficients of the first 50 terms in the binomial expansion of (1-x)¹⁰⁰, is equal to

Solution

Maths sir · 4 steps

Split each ¹⁰⁰Cᵣ with Pascal's rule, and the alternating sum telescopes down to one term.

1. Which coefficients. The general term of (1-x)¹⁰⁰ is ¹⁰⁰Cᵣ(-x)ʳ, so its coefficient is (-1)ʳ ¹⁰⁰Cᵣ. The first 50 terms are r = 0 to r = 49, so we need S = Σᵣ₌₀⁴⁹(-1)ʳ ¹⁰⁰Cᵣ.
2. Split with Pascal. Write ¹⁰⁰Cᵣ = ⁹⁹Cᵣ + ⁹⁹Cᵣ₋₁, taking ⁹⁹C₋₁ = 0. Then every ⁹⁹C term appears twice with opposite signs, except the very last one.

   S = (⁹⁹C₀) - (⁹⁹C₁ + ⁹⁹C₀) + (⁹⁹C₂ + ⁹⁹C₁) - … - (⁹⁹C₄₉ + ⁹⁹C₄₈)

3. Telescope. Everything cancels in pairs and only ⁹⁹C₄₉ survives. It carries the sign of the r = 49 term, (-1)⁴⁹ = -1.
4. So. The sum is -⁹⁹C₄₉, which is option (c).

The idea to keep. For a partial alternating sum of binomial coefficients, use ⁿCᵣ = ⁿ⁻¹Cᵣ + ⁿ⁻¹Cᵣ₋₁: it telescopes to Σᵣ₌₀ᵏ(-1)ʳ ⁿCᵣ = (-1)ᵏ ⁿ⁻¹Cₖ.

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