救命啊…高中数列问题…高手进

来源：百度知道 编辑：UC知道 时间：2024/06/17 10:16:07
设Sn是等差数列{an}前n项和，若以点O(0,0),A(l,Sl),B(m,Sm),C(p,Sp)为顶点的四边形(其中l<m<p)中，AB//OC,则l,m,p之间的等量关系式经化简后为什么？…我要过程…正确追分…谢谢！

向量AB=(m-l,Sm-Sl),向量OC=(p,Sp)
因为AB//OC
故(m-1)/p=(Sm-Sl)/Sp
Sm=ma1+[m(m-1)d]/2,Sl=la1+[l(l-1)d]/2,Sp=pa1+[p(p-1)d]/2
将这三项代入上式可得：
（m-l)/p=[(m-l)a1+(m-l)(m+l-1)d/2]/[pa1+p(p-1)d/2]
约分后得[a1+(m+l-1)d/2]/[a1+(p-1)d/2]=1
将分母乘到右边得a1+(m+l-1)d/2=a1+(p-1)d/2
最终得m+l=p

a(n) = a + (n-1)d, n = 1,2,...

S(n) = na + n(n-1)d/2, n=1,2,...

向量OC = [p, S(p)] = [p, pa + p(p-1)d/2]
向量AB = [m-l,S(m) - S(l)] = [m-l, (m-l)a + [m(m-1) - l(l-1)]d/2]
= [m-l, (m-l)a + [m^2 - l^2 -(m-l) ]d/2]
= [(m-l), (m-l)a + (m-l)(m+l -1)d/2]

AB//OC

p/(m-l) = [pa + p(p-1)d/2]/[(m-l)a + (m-l)(m+l -1)d/2],

a + (m+l - 1)d/2 = a + (p-1)d/2,
m+l - 1 = p-1,
m+l = p.