Question

若数列{an}为等差数列，它的前n项和为Sn，有如下结论 Sm=Sn,(m不等于n),Sn+m=0.?

Sm=a,Sn=b,（m不等于n),Sm+n=-(m+n)?
S偶—S奇=nd,S奇比S偶=an比an+1?

Answer 1

Sm=ma1+m(m-1)d/2
Sn=na1+n(n-1)d/2
Sm=Sn 相减得
(m-n)a1+(m^2-m-n^2+n)d/2=0
m≠n a1+(m+n-1)d/2=0
S(m+n)=(m+n)a1+(m+n)(m+n-1)d/2=(m+n)[a1+(m+n-1)d/2]=0

Sm=n Sn=m
Sn=na1+n(n-1)d/2=m
Sm=ma1+m(m-1)d/2=n 相减
(n-m)a1+(n^2-n-m^2+m)d/2=m-n m≠n
a1+(n+n-1)d/2=-1
S(m+n)=)=(m+n)a1+(m+n)(m+n-1)d/2=(m+n)[a1+(m+n-1)d/2]=-（m+n）

等差数列项数为2n
S奇=a1+a3+a5+……+a(2n-1) (1)
S偶=a2+a4+a6+……+a2n (2)
(2)-(1)
S偶-S奇=(a2-a1)+(a4-a3)+(a6-a5)+……+(a2n-a(2n-1))=nd
S奇=na1+n(n-1)2d/2
S偶=na2+n(n-1)2d/2
S奇/S偶=(a1+(n-1)d)/(a2+(n-1)d)
==(a1+(n-1)d)/(a1+d+(n-1)d)
==(a1+(n-1)d)/(a1+nd)
=an/a(n+1)