NVAMediumJEE Main 2025 · 3 April, Shift 2Simple Applications

Mathematics Question from JEE Main 2025 · 3 April, Shift 2

Let (1+x+x2)10=a0+a1x+a2x2++a20x20(1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \dots + a_{20} x^{20}. If (a1+a3+a5++a19)11a2=121k(a_1 + a_3 + a_5 + \dots + a_{19}) - 11a_2 = 121k, then kk is equal to _____

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Simple Applications questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions