在一个等差数列中,若am+an=ap+aq,如何证明Sm+Sn=Sp+Sq.

ak= a1+(k-1)d
am+an = ap+aq
2a1+ (m+n-2)d = 2a1+(p+q-2)d
m+n= p+q
Sm + Sn
= [(2a1+(m-1)d )m + (2a1+(n-1)d )n ] /2
=[ 2a1(m+n) + ( m^2+n^2 -m-n)d ] /2
={ 2a1(m+n) + [(m+n)^2- 3(m+n)]d } /2
Sp + Sq
= [(2a1+(p-1)d )p + (2a1+(q-1)d )q ] /2
=[ 2a1(p+q) + ( p^2+q^2 -p-q)d ] /2
={ 2a1(p+q) + [(p+q)^2- 3(p+q)]d } /2
={ 2a1(m+n) + [(m+n)^2- 3(m+n)]d } /2
= Sm + Sn