设第一行的公差为d1,第二行的公差为d2,...第n行的公差为dn设公比为qA11 A12=A11+d1A12=qA11 A22=A12+d2=qA11+d2又A22=qA12=q(A11+d1)=qA11+qd1所以d2=qd1同理可得d3=q^2d1,dn=q^(n-1)d1,Ann=q^(n-1)A11+q^(n-1)d1A42=1/8,A43=3/16,得d4=3/16-1/8=1/16A44=3/16+1/16=1/4A44=q^2A24,所以q^2=A44/A24=1/4,q=1/2所以d4=q^3d1=(1/2)^3d1=1/16得d1=1/2d2=qd1=(1/2)*(1/2)=1/4A24=a21+3d2,所以A21=A24-3d2=1-3/4=1/4所以A11=A21/q=(1/4)/(1/2)=1/2Ann=q^(n-1)A11+q^(n-1)d1=(1/2)^(n-1)*(1/2)+(1/2)^(n-1)*(1/2)=(1/2)^(n-1)所以A11+A22+...+Ann=(1-(1/2)^n)/(1-(1/2)=2(1-(1/2)^n)