Sp=(a1+ap)p/2=qp/2*[2a1+(p-1)d]=q pa1+p(p-1)/2*d=q同理qa1+q(q-1)/2*d=p相减(q-p)a1+(q-p)(q+p+1)/2*d=p-q所以a1+(q+p-1)/2*d=-1S(q+p)=[a1+a(p+q)](p+q)/2=[a1+(q+p-1)/2*d](p+q)=-(p+q)=-p-q