Question

在数列{an}中,a1=1/2,an=1-1/an-1(n≥2,n∈N*),数列{an}的前n项和为Sn 求证an+3=a

Answer 1

a1=1/2;
a2=-1;
a3=2;
a4=1/2;
所以，an+3=an;
s(3*k)=(3*k)/2,k>=1;
s(3*k+1)=(3*k+1)/2,k>=0;
s(3*k+2)=(3*k-1)/2,k>=0;