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}中，若Sp=q，Sq=p（p不等于q），则Sp+q=-（p+q） 为什么~

设公差为d 根据求和公式 Sp=pa1+p（p-1）d/2=q Sq=qa1+q（q-1）d/2=p
qSp- pSq=pq（p-q）d/2=q^2-p^2=-（p+q）*（p-q） 故得pqd=-2（p+q）


又 Sp+q=（p+q）a1+（p+q）（p+q-1）d/2=（p+q）a1+(（p+q）^2--（p+q）)d/2
=(pa1+p（p-1）d/2)+qa1+q（q-1）d/2)+pqd=q+p+pqd=q+p-2（p+q）==-（p+q）

解答：解；设等差数列{an}中，首项为a1，公差为d，则Sp=pa1+p(p?1)d2=q，Sq=qa1+q(q?1)d2=p∴d=?2(p+q)qp设p＜q，则Sp+q=Sp+ap+1+ap+2+…+ap+q=Sp+Sq+pqd=p+q+pq?2(p+q)qp=-（p+q）故答案为-（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)

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安康市19810195893： 已知等差数列{an}满足Sp=q,Sq=p求证Sp+q= - (p+q),其中(p≠q) - ？
颜妻曲莱：[答案] 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/2p=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...

安康市19810195893： 已知等差数列{an}满足Sp=q,Sq=p求证Sp+q= - (p+q),其中(p≠q) - ？
颜妻曲莱： 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) 如果不懂,请Hi我,祝学习愉快!

安康市19810195893： 已知等差数列{an}满足:Sp=q,Sq=p,求Sp+q(p不等于q) 要过程!？
颜妻曲莱： Sp=pa1+p(p-1)d/2=q Sq=qa1+q(q-1)d/2=p l两式相减,(p-q)a1+(p-q)(p+q-1)d/2=q-p a1+(p+q-1)d/2=-1 Sp+q=(p+q)a1+(p+q)(p+q-1)d/2 =-(p+q)

安康市19810195893： 一道等差数列的题..若等差数列{an}中,Sp=q,Sq=p,则Sp+q=? - ？
颜妻曲莱：[答案] 由公式Sn=na1+n(n-1)d/2有 Sp=pa1+p(p-1)d/2=q.(1) Sq=qa1+q(q-1)d/2=p.(2) (1)-(2)得(p-q)a1+(p+q-1)(p-q)d/2=q-p ∵p≠q ∴p-q≠0 ∴a1+(p+q-1)d/2=-1 ∴S(p+q)=(p+q)a1+(p+q)(p+q-1)d/2=(p+q)[a1+(p+q-1)d/2]=-(p+q)

安康市19810195893： {An}为等差数列,知道了Sp=q和Sq=p 证明一下Sp+q= - p - qSp是前P项和~不太会用公式~ - ？
颜妻曲莱：[答案] 因为他是等差数列,所以前n项和可以写成一个二次函数的形式记为Sn=an^2+bn,则Sp=ap^2+bp=q,Sq=aq^2+q=p,则Sp-Sq=a(p^2-q^2)+b(p-q)=q-p,因为pq不相等,所以a(p+q)+b=-1,两边同乘以p+q得a(p+q)^2+b(p+q)=-p-q= S(p+q)

安康市19810195893： 在等差数列{an}中,已知Sp=q,Sq=p,(p≠q),则Sp+q=______. - ？
颜妻曲莱：[答案] 解;设等差数列{an}中,首项为a1,公差为d,则Sp=pa1+ p(p−1)d 2=q,Sq=qa1+ q(q−1)d 2=p ∴d= −2(p+q) qp 设p
安康市19810195893： 在等差数列an中,已知Sp=q,Sq=p,求Sp+q的值 - ？
颜妻曲莱： Sp=p*a(p/2+1/2)=qa(p/2+1/2)=q/p 同理 a(q/2+1/2)=p/q d=[a(q/2+1/2)-a(p/2+1/2)]/[(q/2+1/2)-(p/2+1/2)] =(p/q-q/p)/(q/2-p/2) =(p^2-q^2)[p*q*(q-p)/2] =-2(p+q)/(p*q) a(p/2+q/2+1/2)=a(p/2+1/2)+d*q/2 =q/p-(q/2)*2(p+q)/(p*q) =q/p-(p+q)/p =-1 S(p+q)=(p+q)*a(p/2+q/2+1/2) =(p+q)*(-1) =-(p+q)

安康市19810195893： 在等差数列An中,已知Sp=q,Sq=p,(p不等于q),求Sp+q的值. - ？
颜妻曲莱： 设等差数列首项为a1,公差为d,根据题意: sp=(a1+ap)p/2=q,即:a1+a1+(p-1)d=2q/p,所以: 2a1+(p-1)d=2q/p.....(1) sq=(a1+aq)q/2=p,即:a1+a1+(q-1)d=2p/q,所以: 2a1+(q-1)d=2p/q.....(2) 根据(1),(2)可到: a1=(q^2+p^2+pq-p-q)/pq. d=-2(p+q)/pq; 所以: sp+q=(a1+ap+q)(p+q)/2=-(p+q).

安康市19810195893： 在等差数列{an}中,Sp=q,Sq=p,Sp+q的值 - ？
颜妻曲莱： s=a1+(n-1)d 根据Sp=q,Sq=p 得:q=a1+(p-1)dp=a1+(q-1)d 两式联立得d=-1 a1=p+q-1 带入下式即可Sp+q=a1+(p+q)d = -1

安康市19810195893： 已知等差数列an的前n项和为sn,且sp=q,sq=p,(p、q∈N*,p≠q) - ？
颜妻曲莱： Sp=pa1+p(p-1)d/2=q pqa1+pq(p-1)d/2=q^2 Sq=qa1+q(q-1)d/2 =p pqa1+pq(q-1)d/2=p^2 相减 p*q*d/2*(p-q)=(q^2-p^2) p*q*d/2=p+q d=2(p+q)/pq