作业帮已知等差数列{an}满足Sp=q,Sq=p求证Sp+q=-(p+q),其中（p≠q）
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已知等差数列{an}满足Sp=q,Sq=p求证Sp+q=-(p+q),其中（p≠q）
已知等差数列{an}满足Sp=q,Sq=p求证Sp+q=-(p+q),其中（p≠q）
已知等差数列{an}满足Sp=q,Sq=p求证Sp+q=-(p+q),其中（p≠q）
等差数列{an}的前n项和为Sn,Sp=q,Sq=p,p≠q,则S(p+q)=-(p+q)
证明：由题意,
q=Sp=a1+a2+...+ap=pa1+p(p-1)d/2
p=Sq=a1+a2+...+aq=qa1+q(q-1)d/2
两式相减,得到
q-p=(p-q)[a1+(p+q-1)d/2]
因为p≠q,故
a1+(p+q-1)d/2=-1
因此
S(p+q)=a1+a2+...+a(p+q)=(p+q)(a1+a(p+q))/2
=(p+q)(a1+a1+(p+q-1)d)/2
=(p+q)(a1+(p+q-1)d/2)
=(p+q)*(-1)
=-(p+q)
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因为是等差数列，所以
可以设sn=an^2+bn
所以
sp=ap^2+bp=q
sq=aq^2+bq=p
解得
a=-(p+q)/pq,b=(p^2+pq+q^2)/pq
s(p+q)=a(p+q)^2+b(p+q)
=-(p+q)/pq*(p+q)^2+(p^2+pq+q^2)/pq*(p+q)
=(p+q)[(p^2+pq+q^2)-(p+q)^2]/pq
=(p+q)[-pq]/pq
=-(p+q)