作业帮证明：若等差数列S(p)=q,S(q)=p,则S(p+q)=-(p+q)
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证明：若等差数列S(p)=q,S(q)=p,则S(p+q)=-(p+q)
证明：若等差数列S(p)=q,S(q)=p,则S(p+q)=-(p+q)
证明：若等差数列S(p)=q,S(q)=p,则S(p+q)=-(p+q)
等差数列前n项和S=an^2+bn,
∴S/n=an+b,
∴[S
/p-S/q]/(p-q)=[S/(p+q)-S
/p]/[p+q-p],
即[q/p-p/q]/(p-q)=[S/((p+q)-q/p]/q,
∴-（p+q)/p=S/(p+q)-q/p,
∴S=-(p+q).
详细过程a(n)=a+(n-1)d,
s(n)=na+n(n-1)d/2,
q=s(p)=pa+p(p-1)d/2, q/p=a+(p-1)d/2,
p=s(q)=qa+q(q-1)d/2, p/q=a+(q-1)d/2.
q/p-p/q = (p-q)d/2 = (q^2-p^2)/(pq), d/2 = -(p+q)/(pq).
a=q/p - (p-1...
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详细过程a(n)=a+(n-1)d,
s(n)=na+n(n-1)d/2,
q=s(p)=pa+p(p-1)d/2, q/p=a+(p-1)d/2,
p=s(q)=qa+q(q-1)d/2, p/q=a+(q-1)d/2.
q/p-p/q = (p-q)d/2 = (q^2-p^2)/(pq), d/2 = -(p+q)/(pq).
a=q/p - (p-1)d/2 = q/p + (p-1)(p+q)/(pq) = [p^2-p+pq-q+q^2]/(pq)
s(p+q)=(p+q)a+(p+q)(p+q-1)d/2=(p+q)[p^2-p+pq-q+q^2]/(pq) + (p+q)(p+q-1)[-(p+q)/(pq)]
=(p+q)/(pq)[p^2-p+pq-q+q^2 - (p+q)(p+q-1)]
=(p+q)/(pq)[p^2-p+pq-q+q^2 - p^2-q^2-2pa + p+q]
=(p+q)/(pq)[-pq]
=-(p+q)
收起
a(n)=a+(n-1)d,
s(n)=na+n(n-1)d/2,
q=s(p)=pa+p(p-1)d/2, q/p=a+(p-1)d/2,
p=s(q)=qa+q(q-1)d/2, p/q=a+(q-1)d/2.
q/p-p/q = (p-q)d/2 = (q^2-p^2)/(pq), d/2 = -(p+q)/(pq).
a=q/p - (p-1)d/2...
全部展开
a(n)=a+(n-1)d,
s(n)=na+n(n-1)d/2,
q=s(p)=pa+p(p-1)d/2, q/p=a+(p-1)d/2,
p=s(q)=qa+q(q-1)d/2, p/q=a+(q-1)d/2.
q/p-p/q = (p-q)d/2 = (q^2-p^2)/(pq), d/2 = -(p+q)/(pq).
a=q/p - (p-1)d/2 = q/p + (p-1)(p+q)/(pq) = [p^2-p+pq-q+q^2]/(pq)
s(p+q)=(p+q)a+(p+q)(p+q-1)d/2=(p+q)[p^2-p+pq-q+q^2]/(pq) + (p+q)(p+q-1)[-(p+q)/(pq)]
=(p+q)/(pq)[p^2-p+pq-q+q^2 - (p+q)(p+q-1)]
=(p+q)/(pq)[p^2-p+pq-q+q^2 - p^2-q^2-2pa + p+q]
=(p+q)/(pq)[-pq]
=-(p+q)
收起