MathematicsMedium64×since 2002Q2910For x ∈\in∈ R, x ≠\ne= -1, if (1 + x)²⁰¹⁶ + x(1 + x)²⁰¹⁵ + x²(1 + x)²⁰¹⁴ + . . . . + x²⁰¹⁶ = ∑i=02016ai xi, \sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,i=0∑2016aixi, then a₁₇ is equal to :A2017!17! 2000!{{2017!} \over {17!\,\,\,2000!}}17!2000!2017!B2016!17! 1999!{{2016!} \over {17!\,\,\,1999!}}17!1999!2016!C2017!2000!{{2017!} \over {2000!}}2000!2017!D2016!16!{{2016!} \over {16!}}16!2016!Check answerSkip