等差数列中na1+n(n-1)d/2=-dn²/2+(a1+d/2)n,
∴可设Sn=A n²+Bn.
1.（Sp-Sq）/(p-q)=( A p²+Bp- A q²-Bq) / (p-q)
=[( A p² - A q²)+(Bp -Bq) ]/ (p-q)=A(p+q)+B.
S(p+q)/(p+q)= ( A (p+q)²+B(p+q) )/ (p+q) =A(p+q)+B.
∴（Sp-Sq）/(p-q)= S(p+q)/(p+q).
2.Sm=Sn,则A m²+Bm =A n²+Bn.
A(m+n)(m-n)+B(m-n)=0,
A(m+n) +B=0,
∴S(m+n)= A (m+n)²+B(m+n)=( A(m+n) +B)(m+n)=0.