已知等差数列an的前n项和为sn,且sm=sn(m不等于n)求s(m+n)

假设m>nSn=A1+A2+……+AnSm=A1+A2+……+An+A(n+1)+A(n+2)+……+AmSm-Sn=A(n+1)+A(n+2)+……+Am=0 (共m-n项)从A(n+1)项到Am项也是等差数列A(n+1)+A(n+2)+……+Am=(A(n+1)+Am)(m-n)/2=0A(n+1)+Am=0A(n+1)+Am=An+A(m+1)=A(n-1)+A(m+2)=……=A1+A(m+n)=0S(m+n)=(A1+A(m+n))(m+n)/2=0