Given that the sum of the coefficients of the first k terms in the binomial expansion of (1-x)¹⁰⁰ is equal to ⁹⁹C₅₀ , find the value of k .

Step-by-step solution

Let the sum of the first k terms be S_k . The binomial expansion of (1-x)¹⁰⁰ is: (1-x)¹⁰⁰ = ¹⁰⁰C₀ - ¹⁰⁰C₁x + ¹⁰⁰C₂x^2 - + ¹⁰⁰C₁₀₀x¹⁰⁰ The sum of the coefficients of the first k terms is: S_k = _ r=0 ^ k-1 (-1)^r ¹⁰⁰C_ r Using Pascal's identity, ¹⁰⁰C_ r = ⁹⁹C_ r + ⁹⁹C_ r-1 (with ⁹⁹C_ -1 = 0 ). S_k = ⁹⁹C₀ - (⁹⁹C₁ + ⁹⁹C₀) + (⁹⁹C₂ + ⁹⁹C₁) - + (-1)^ k-1 (⁹⁹C_ k-1 + ⁹⁹C_ k-2 ) This is a telescoping sum. All intermediate terms cancel out, leaving: S_k = (-1)^ k-1 ⁹⁹C_ k-1 We are given that S_k = ⁹⁹C₅₀ . Therefore, (-1)^ k