问题
填空题
已知数列{an}的前n项和Sn=3-3×2n,则an=______.
答案
①当n=1时,a1=S1=3-3×21=-3;
②当n≥2时,an=Sn-Sn-1=(3-3×2n)-(3-3×2n-1)=-3×2n-1;
综合①②,得an=-3×2n-1(n∈N*).
故答案为:-3×2n-1(n∈N*).
已知数列{an}的前n项和Sn=3-3×2n,则an=______.
①当n=1时,a1=S1=3-3×21=-3;
②当n≥2时,an=Sn-Sn-1=(3-3×2n)-(3-3×2n-1)=-3×2n-1;
综合①②,得an=-3×2n-1(n∈N*).
故答案为:-3×2n-1(n∈N*).