JEE Main 2026 Mathematics question
Let the smallest value of k ∈ ℕ, for which the coefficient of x³ in (1+x)³ + (1+x)⁴ + (1+x)⁵ + … + (1+x)⁹⁹ + (1+kx)¹⁰⁰, x ≠ 0, is for some n ∈ ℕ, be p. Then the value of p + n is:
Answer: (B)
Solution
By the hockey-stick identity, Σⱼ₌₃⁹⁹C(j,3) = C(100,4), and the coefficient from (1+kx)¹⁰⁰ is k³C(100,3). Using C(100,4)/C(100,3)
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