Question

等差数列中，若Sm=Sp（m≠p），则Sm+p=0怎么证明

Answer 1

等差数列前n项和公式为Sn=an^2+bn ，图像为过原点的二次曲线，你画任意一个图像【只有右半部分x>0}，上面取Sn值相等的两点，你会发现它们关于对称轴对称，且m+p为该图像一个零点，即Sm+p=0

追问

那么若Sm=p,Sp=m（m≠p）,则Sm+p=-（m+p）是为什么

Answer 2

a(n)=a+(n-1)d,
s(n)=na+n(n-1)d/2,

s(m)=ma+m(m-1)d/2 = s(p) = pa + p(p-1)d/2,
0=(p-m)a + d/2[p^2 - p - m^2 +m] = (p-m)a + [(p-m)(p+m) - (p-m)]d/2
=(p-m)[a + (p+m-1)d/2],
m≠p,
0 = a + (p+m-1)d/2,

s(m+p)=(m+p)a + (m+p)(m+p-1)d/2 = (m+p)[a + (m+p-1)d/2] = (m+p)*0 = 0.

-----------------------------
s(m)=p, s(p)=m, m≠p,
p = s(m) = ma + m(m-1)d/2,
m = s(p) = pa + p(p-1)d/2,

m-p = (p-m)a + d/2[p^2 - p - m^2 + m] = (p-m)a + [(p-m)(p+m) - (p-m)]d/2,
= (p-m) [ a + (p+m-1)d/2],
m≠p,
-1 = a + (p+m-1)d/2,
s(m+p) = (m+p)a + (m+p)(m+p-1)d/2
=(m+p)[a + (m+p-1)d/2] = (m+p)[-1] = -(m+p).