1.在等差数列{an}中,已知Sn=m,Sm=n,（n不等于m）,求S（m+n）?

设首项为a1,公差为d,Sn=na1+n*(n-1)d/2=m,
Sm=ma1+m*(m-1)d/2=n,
两式相减,得(n-m)a1+[(n-m)(n+m)-(n-m)]d/2=-(n-m)
a1+[n+m-1]d/2=-1
S(m+n)=(n+m)a1+[(n+m)(n+m-1)]d/2=(n+m)[a1+[n+m-1]d/2]=-(n+m)