等差数列前n项和为Sn ,若Sn=Sm(m>

Sn=na1＋(1/2)n(n＋1)d
Sm=ma1＋(1/2)m(m＋1)d
两式相减,得：
0=(n－m)a1＋(1/2)d[(n²－m²)＋(n－m)] 两边除以n－m,得：
a1＋(1/2)(m＋n＋1)d=0
而：S(m＋n)=(m＋n)a1＋(1/2)(m＋n)(m＋n＋1)d
=(m＋n)[a1＋(1/2)(m＋n＋1)d]
=(m＋n)×0
=0