等差数列an前n项和为Sn=m,Sm=n,求Sm+n的值

设公差为d.
Sm=ma1+(m^2-m)d/2=n,则mna1+(m^2n-mn)d/2=n^2 (1)
Sn=na1+(n^2-n)d/2=m,则mna1+(mn^2-mn)d/2=m^2 (2)
(1)-(2)得：(d/2)(m^2n-mn^2)=n^2-m^2、(d/2)mn(m-n)=-(m+n)(m-n)
因为mn,所以(d/2)mn=-(m+n)、mnd=-2m-2n (*)
S(m+n)=a1+a2+…+am+a(m+1)+a(m+2)+…+a(m+n)
=Sm+[a1+md]+[a1+(m+1)d]+…+[a1+(m+n-1)d]
=n+{a1+(a1+d)+…+[a1+(n-1)d]}+mnd
=n+[a1+a2+…+an]+mnd
=n+Sn+mnd
=m+n+mnd
将(*)式代入可得：S(m+n)=-m-n