在等差数列已知数列{an} 中,设前m项和为Sm,前n项和为Sn,且Sm=Sn (m不等于n）,求Sm+n

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