Question

等差数列 Sn=m Sm=n S（n+m）=-（m+n）推导过程

Answer 1

sn=a1n+n(n-1)d/2=m
sm=a1m+m(m-1)d/2=n
相减。
a1=(m-n(n-1)d/2)/n
a1=(n-m(m-1)d/2)/m
相除：(m-n(n-1)d/2)/n=(n-m(m-1)d/2)/m
m^2-mn(n-1)d/2=n^2-mn(m-1)d/2
d/2(m^2n-mn-mn^2+mn)=n^2-m^2
d=2(n^2-m^2)/(m^2n-mn^2)
d=2(n-m)(n+m)/(mn(m-n)
d=-2(n+m)/(mn).........................1
s(n+m)=a1(n+m)+(n+m)(n+m-1)d/2
=a1n+a1m+(n^2+2mn+m^2-n-m)d/2
=m-(n^2-n)d/2+n-(m^2-m)d/2+(n^2+2mn+m^2-n-m)d/2
=m+n-(n^2-n+m^2-m-n^2-2mn-m^2+n+m)d/2
=m+n-(2mn)d/2
=m+n-mnd/2
将1式代入：
S(n+m)=m+n-mn*(-2(n+m))/(mn)=m+n-2(m+n)=-(m+n)

Answer 2

设一下步姐姐了