(1)
依题意得:Xi=Yi^2 ,X(i+1)=Y(i+1)^2,
由(Xi,Yi)确定的直线方程为:X/Xi+Y/Yi=1
亦知X(i+1)/Xi+Y(i+1)/Yi=1 将Xi=Yi^2 ,X(i+1)=Y(i+1)^2代入得:
Y(i+1)^2/Yi^2+Y(i+1)/Yi=1 即Y(i+1)^2/Yi^2+Y(i+1)/Yi-1=0
对比一元二次方程ax^2+bx+c=0形式知 Y(i+1)/Yi为方程Y(i+1)^2/Yi^2+Y(i+1)/Yi-1=0的根为一常数,即Y(i+1)/Yi=t(t为常数),因而{yi}是等比数列;
(2)
算出上题Y(i+1)/Yi=t为t=(√5-1)/2
Si=Xi*Yi=Yi^3
lim(S1+S2+……+Sn)
=lim(Y1^3+Y2^3+……+Yn^3)
=lim(m^3+(m*t)^3+(m*t^2)^3++……+(m*t^(n-1))^3
=lim(m^3*(1-t^3n)/(1-t^3))
=lim(m^3*(1-((√5-1)/2)^3n)/(1-((√5-1)/2)^3))
=((3+√5)/4)*m^3