Search for: if (1+x)15=C0+C1x+C2x2+…+C15x15 then C2+2C3+3C4+…+14C15 is equal to if (1+x)15=C0+C1x+C2x2+…+C15x15 then C2+2C3+3C4+…+14C15 is equal to A14⋅214B13⋅214+1C13⋅214−1DNone of these Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required) I agree to the terms and conditions and privacy policy. Solution:we have (1+x)15=C0+C1x+C2x2+…+C15x15⇒ (1+x)15−1x=C1+C2x+…+C15x14On differentiating both sides w.r.t. x, we getx⋅15(1+x)14−(1+x)15+1x2=C2+2C3x+…+14C15x13On putting x = t, we get C2+2C3+…+14C15=15⋅214−215+1=13⋅214+1Post navigationPrevious: if a and d, are two complex numbers, then the sum to (n+ 1) terms of the following series aC0−(a+d)C1+(a+2d)C2−…+… is Next: if 1+x+x2n=∑r=02n arxr, then a1−2a2+3a3…−2na2n is equal toRelated content JEE Main 2023 Question Papers with Solutions JEE Main 2024 Syllabus Best Books for JEE Main 2024 JEE Advanced 2024: Exam date, Syllabus, Eligibility Criteria JEE Main 2024: Exam dates, Syllabus, Eligibility Criteria JEE 2024: Exam Date, Syllabus, Eligibility Criteria NCERT Solutions For Class 6 Maths Data Handling Exercise 9.3 JEE Crash Course – JEE Crash Course 2023 NEET Crash Course – NEET Crash Course 2023 JEE Advanced Crash Course – JEE Advanced Crash Course 2023