在等差数列{a}中前n项和为Sn,若Sm=p,Sp=m则Sm+p=-(m+P)如何证明

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在等差数列{a}中前n项和为Sn,若Sm=p,Sp=m则Sm+p=-(m+P)如何证明
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在等差数列{a}中前n项和为Sn,若Sm=p,Sp=m则Sm+p=-(m+P)如何证明
在等差数列{a}中前n项和为Sn,若Sm=p,Sp=m则Sm+p=-(m+P)如何证明

在等差数列{a}中前n项和为Sn,若Sm=p,Sp=m则Sm+p=-(m+P)如何证明
证明:由数列为等差数列,可设其前n项和Sn=An^2+Bn
Sm=Am^2+Bm=p,(1)
Sp=Ap^2+Bp=m (2)
(1)+(2)得A(m^2+p^2)+B(m+p)=m+p
p* (1) -m*(2) 整理得 mpA=-(m+p)
所以Sm+p =A(m+p)^2 +B(m+p)
=A(m^2+2mp+p^2)+B(m+p)
=A(m^2+p^2)+B(m+p)+2mpA
=m+p-2(m+p)
=-(m+p)

证:
设公差为d
Sm=Sp
ma1+m(m-1)d/2=pa1+p(p-1)d/2
(m-p)a1+[m(m-1)-p(p-1)]d/2=0
(m-p)a1+[(m²-p²)-(m-p)]d/2=0
(m-p)a1+[(m+p)(m-p)-(m-p)]d/2=0
(m-p)a1+(m-p)(m+p-1)d/2=0
m≠p,等式两边同乘以(m+p)/(m-p)
(m+p)a1+(m+p)(m+p-1)d/2=0
S(m+p)=(m+p)a1+(m+p)(m+p-1)d/2=0