Soit la suite un=(−4n+2)(2n+n2)u_n = (-4n +2)(2n+n^2)un=(−4n+2)(2n+n2) pour n=0,1,2,…n=0,1,2,\dotsn=0,1,2,… Calculer u0u_0u0 et u1u_1u1.
u0=0u_0 = 0u0=0 et u1=6u_1 = 6u1=6
u0=0u_0 = 0u0=0 et u1=−6u_1 = -6u1=−6
u0=0u_0 = 0u0=0 et u1=3u_1 = 3u1=3